MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 3 - MCQExams.com
CBSE
Class 11 Engineering Physics
Motion In A Straight Line
Quiz 3
A ball is projected vertically up with a velocity of $$40$$ $$m/s$$. At the same time another ball is dropped from a height of $$100$$ m. The magnitudes of their velocities are equal after a time of $$(g= 10m/s^{2})$$
Report Question
0%
2 sec
0%
1 sec
0%
3 sec
0%
4 sec
Explanation
$$v_A = 40 - gt$$
$$v_B = gt$$
$$|v_A| = |v_B|$$
$$\Rightarrow 40 - gt = gt$$
$$t = \dfrac {40}{2 \times 10} = 2\ sec$$
A car, starting from rest, accelerates at the rate $$f$$ through a distance $$S$$, then continues at constant speed for time $$t$$ and then decelerate at the rate $$\frac{f}{2}$$ to come to rest. If the total distance travelled is $$15S$$, then
Report Question
0%
$$S=ft$$
0%
$$S=\dfrac{1}{6}ft^{2}$$
0%
$$S=\dfrac{1}{72}ft^{2}$$
0%
$$S=\dfrac{1}{4}ft^{2}$$
Explanation
$$\displaystyle S_1 = \dfrac{1}{2}ft_{1}^{2}=S$$ since given $$S_1 = S$$
$$S_2 =( ft_1)t_2$$
Since acceleration for $$t_3$$ is $$\dfrac{-f}{2}$$, $$t_3=2t_1$$
$$S_3 =( ft_1)t_3 - \dfrac{1}{2}(\dfrac{f}{2})t_3 ^2$$
$$ \therefore S_3 = ( ft_1)(2t_1) - \dfrac{1}{2}(\dfrac{f}{2})(2t_1)^2= 2ft_1^2=2S_1=2S$$
Given, $$S_1 + S_2 +S_3 = 15S$$
$$\Rightarrow S + S_2 + 2S =15S$$
$$\Rightarrow S_2 = 12S$$
$$\Rightarrow ft_1t_2 = 12S= 12 \times \dfrac{1}{2} \times ft_1^2$$
$$ \Rightarrow t_2 = 6t_1$$
$$S =\dfrac{1}{2}ft_1 ^2 = \dfrac{1}{72}ft_2 ^2 = \dfrac{1}{72}ft ^2$$
(since given $$t_2 = t$$)
A stone is dropped from a height $$h$$. Simultaneously another stone is thrown up from the ground which reaches the height $$4h$$. The two stones cross each other after a time:-
Report Question
0%
$$\sqrt{\dfrac{h}{2g}}$$
0%
$$\sqrt{\dfrac{h}{8g}}$$
0%
$$\sqrt{8hg}$$
0%
$$\sqrt{2hg}$$
Explanation
For B: $$u$$ is such that stone B reaches uh.
0 at $$max^m$$ pt
Hence, $$ v^2 = u^2 - 2g_B$$
$$u^2 = 2g (uh) \Rightarrow u = \sqrt {8gh} \ \ \ .....(1)$$
Now $$eq^n $$ for B: $$S_B = ut - \dfrac {1}{2} gt^2 \ \ \ ..... (2)$$
$$eq^n$$ for A: $$S_A = \dfrac {1}{2} gt^2 \ \ \ ..... (3)$$
$$S_A + S_B = h = ut \Rightarrow t = \dfrac {h}{u}$$
From (1), $$u = \sqrt {8gh} \Rightarrow t = \sqrt {\dfrac {h}{8g}}$$
A body is thrown vertically upwards with an initial velocity $$u$$ reaches a maximum height in $$6s$$. The ratio of the distance travelled by the body in the first second to the seventh second is
Report Question
0%
1:1
0%
11:1
0%
1:2
0%
1:11
Explanation
height in the first second,
$$h_{1}=u-\displaystyle \frac{g}{2}$$
hight covered in the first second of downward journey
$$h_{2}=\displaystyle \frac{g}{2}$$
given $$t_{a}=\displaystyle \frac{u}{g}=6; u=6g$$
$$v = 0 $$ at max height
$$v^2 = u^2 - 2gs | t = \dfrac {u}{g}$$
$$\Rightarrow u^2 = 2gH_{max} | u = gt$$
$$u = \sqrt {2gH_{max}} | u = 6g - (0)$$
$$s_1 = u(1) - \dfrac {1}{2}g(1)^2 = u - \dfrac {g}{2} \ \ \ ..... (1)$$
$$s_7 = [u(7)-\dfrac {1}{2}g(7)^2] - [u(6) - \dfrac {1}{2}g(6)^2]$$
$$= u -\dfrac {1}{2}g(13) \ \ \ ..... (2)$$
$$\dfrac {s_1}{s_7} = \dfrac {u-\dfrac {g}{2}}{u - \dfrac {g}{2}(13)} = \dfrac {6g - \dfrac {g}{2}}{6g - \dfrac {13g}{2}} = \dfrac {11}{-1} \ \ \ ..... (3)$$
So, $$\dfrac {s_1}{s_7} = \dfrac {11}{1}$$
A point moves with uniform acceleration. Let $$v_{1}$$, $$v_{2}$$, $$v_{3}$$ denote the average velocities in three successive intervals of time $$t_{1}$$, $$t_{2}$$, $$t_{3}$$. Correct relation among the following is
Report Question
0%
$$(v_{1}-v_{2})$$ : $$(v_{2}-v_{3})$$ = $$(t_{1}-t_{2})$$ : $$(t_{2}-t_{3})$$
0%
$$(v_{1}-v_{2})$$ : $$(v_{2}-v_{3})$$ = $$(t_{1}+t_{2})$$ : $$(t_{2}+t_{3})$$
0%
$$(v_{1}-v_{2})$$ : $$(v_{2}-v_{3})$$ = $$(t_{1}-t_{2})$$ : $$(t_{2}+t_{3})$$
0%
$$(v_{1}-v_{2})$$ : $$(v_{2}-v_{3})$$ = $$(t_{1}+t_{2})$$ : $$(t_{2}-t_{3})$$
Explanation
$${ v }_{ avg }=\dfrac { ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 } }{ t } =u+\dfrac { 1 }{ 2 } at$$
$${V}_{1}={U}+\dfrac{a{t}_{1}}{2}$$
$${V}_{2}={U}_{1}+\dfrac{a{t}_{2}}{2}$$
$${V}_{3}={U}_{2}+\dfrac{a{t}_{3}}{2}$$
so we can find,
$$({V}_{1}-{V}_{2}):({V}_{2}-{V}_{3})=({t}_{1}-{t}_{2}):({t}_{2}-{t}_{3})$$
Alternatively,
Since acceleration is constant, acceleration in the interval $$t_1 -t_2$$ will be same as acceleration in the interval $$t_2 -t_3$$
Hence,
$$({V}_{1}-{V}_{2}):({V}_{2}-{V}_{3})=({t}_{1}-{t}_{2}):({t}_{2}-{t}_{3})$$
A man in a lift ascending with an upward acceleration throws a ball vertically upwards and catches it after $$t_{1}$$ second. Later when the lift is descending with the same acceleration, the man throws the ball up again with same velocity and
catches it after $$t_{2}$$ second.
A) The acceleration of the elevator is $$g\dfrac{(t_{2}-t_{1})}{(t_{1}+t_{2})}$$
B) The velocity of projection of the ball relative to
elevator is $$\dfrac{t_{2}t_{1} g}{t_{1}+t_{2}}$$.
We can conclude that:
Report Question
0%
only $$A$$ is true
0%
only $$B$$ is true
0%
Both $$A$$ and $$B$$ is true
0%
Both $$A$$ and $$B$$ are false
Explanation
We know that
$$ { t }_{ 1 }=\dfrac { 2u }{ g+a } \quad \& \quad { t }_{ 2 }=\dfrac { 2u }{ g-a }$$
We get,
$$2u={ t }_{ 1 }(g+a)={ t }_{ 2 }(g-a)\\ a({ t }_{ 1 }+{ t }_{ 2 })=g({ t }_{ 2 }-{ t }_{ 2 })\\ a=\dfrac { g({ t }_{ 2 }-{ t }_{ 1 }) }{ { t }_{ 1 }+{ t }_{ 2 } } $$
Now, the velocity of projection of the ball relative to
elevator is u,
We have,
$$g +a = \displaystyle \frac{2u}{t_1}$$
$$g - a = \displaystyle \frac{2u}{t_2}$$
adding both,
$$2g = 2u \displaystyle(\frac{1}{t_1} + \frac{1}{t_1})$$
$$\therefore u = \displaystyle \frac{t_1t_2g}{t_1+t_2}$$
A body is at rest at x =At t = 0, it starts moving in the positive x-direction with a constant acceleration. At the same instant another body passes through x = 0 moving in the positive x-direction with a constant speed. The position of the first body is given by $$x_{1}(t)$$ after time t and that of the second body by $$ x_{2} (t)$$ after the same time interval. Which of the following graphs correctly describes $$( x_{1} - x_{2})$$ as a function of time t ?
Report Question
0%
0%
0%
0%
Explanation
Relative distance at any point
$$S_{1} (t) = 1/2 at^{2}$$
$$S_{2} (t) = Vt$$
$$\Rightarrow (S_{1} - S_{2}) =1/2 at^{2}-vt$$
Curve in D seems to be correct answer.
An object falls from a bridge that is $$45\ m$$ above water. It falls directly into a small boat moving with constant velocity that is $$12\ m$$ from the point of impact when the object was released. The speed of the boat is
Report Question
0%
$$3\;m/s$$
0%
$$4\;m/s$$
0%
$$5\;m/s$$
0%
$$6\;m/s$$
Explanation
Boat has covered $$12\ m$$ in the same time as taken by ball A to fall down, $$t = \dfrac {12}{v}$$
Time taken by A $$\Rightarrow 45 = \dfrac {1}{2}gt^2$$
$$t = \sqrt {\dfrac {90}{g}}=3 sec $$
Now, $$\dfrac {12}{v} = 3 $$
$$\Rightarrow v = 4 \ m/s$$
A body, projected vertically upwards, crosses a point twice its journey at a height $$h$$ just after $$t_{1}$$ and $$t_{2}$$ seconds. Maximum height reached by the body is
Report Question
0%
$$\dfrac{g}{4}(t_{1}+t_{2})^{2}$$
0%
$$g\left [\dfrac{t_{1}+t_{2}}{4}\right ]^{2}$$
0%
$$2g\left [\dfrac{t_{1}+t_{2}}{4}\right ]^{2}$$
0%
$$\dfrac{g}{4} (t_{1}t_{2})$$
Explanation
$$s_A = ut - \dfrac {1}{2}gt^2 (at \;t = t_1, t_2)$$
So, $$s_A = ut_1 - \dfrac {1}{2}gt_1^2 - (1)$$
$$s_A = ut_2 - \dfrac {1}{2} gt_2^2 - (2)$$
Equating $$(1) and (2)$$
$$\Rightarrow u (t_1 - t_2) = \dfrac {1}{2} g (t_1^2 - t_2^2)$$
$$\Rightarrow u = g \dfrac {(t_1 + t_2)}{2}$$
$$Max\ ht = \dfrac {u^2}{2g} = \dfrac {g^2 (t_1 + t_2)^2}{4 \times 2g} = 2g (\dfrac {t_1 + t_2}{4})^2$$
A train is moving forward at a velocity of $$2.0$$ $$m/s$$. At the instant the train begins to accelerate at $$0.80 m/s^{2}$$, a passenger drops a coin which takes $$0.50\ s$$ to fall on the floor. Relative to a spot on the floor directly under the coin at release, the coin lands
Report Question
0%
$$1.1$$ m towards the rear of the train
0%
$$1.0$$ m towards the rear of the train
0%
$$0.10$$ m towards the rear of the train
0%
$$0.90$$ m towards the front of the train
Explanation
We have to solve this problem in reference to frame of train
So, Relative velocity of coin w.r.t train $$= 0 m/s$$
Relative acceleration of coin w.r.t train
= -0.8
$$m/{ { s }^{ 2 } }$$
Here, (-) indicates that direction opposite to train initial direction
Then, by using 2nd equation of motion
$$ S=ut+$$ $$\frac { 1 }{ 2 } a{ t }^{ 2 }$$
$$ S= 0+$$ $$\frac { 1 }{ 2 } \times (-0.8)\times { (0.5) }^{ 2 }$$
$$ S= -0.10 m$$
Therefore, the final answer is the coin moves 0.01 m towards the rear of the train.
A ball is thrown vertically upwards with a velocity $$u$$ from the balloon descending with velocity $$v$$. The ball will pass by the balloon after time
Report Question
0%
$$\dfrac{u-\nu}{2g}$$
0%
$$\dfrac{2(u+\nu)}{g}$$
0%
$$\dfrac{2u-\nu}{g}$$
0%
$$\dfrac{2u+\nu}{g}$$
Explanation
$${ \vec { v } }_{ BB }=$$ Relative velocity of ball with respect to balloon $$=\vec { u } +\vec { v }$$
$$0=-\left( u+v \right) +gt$$ $$\implies t=\displaystyle\dfrac { u+v }{ g }$$
Total time $$=\displaystyle\dfrac { 2\left( u+v \right) }{ g }$$
An elevator ascends with an upward acceleration of $$0.2\ m/s^{2}$$. At the instant when its upward speed is $$2\ m/s$$, a loose bolt $$5\ m$$ high from the floor drops from the ceiling of the elevator. The time taken by the bolt to strike the floor and the distance it has fallen are (Take $$g=9.8\ m/s^2$$)
Report Question
0%
$$1 s$$, $$1.9 m$$
0%
$$1 s$$, $$2.9 m$$
0%
$$1 s$$, $$4.9 m$$
0%
$$1 s$$, $$3.9 m$$
Explanation
$$g=9.8\quad \quad a=0.2\quad \\ { a }_{ rel }=10\quad \\ Now,\quad s=\dfrac { 1 }{ 2 } { a }_{ rel }{ t }^{ 2 }\\ t=\sqrt { \dfrac { 2\times 5 }{ 10 } } =1\quad second.\\ Distance\quad covered\quad by\quad elevator=\dfrac { 1 }{ 2 } \times 0.2\times 1\\ \therefore \quad Displacement\quad of\quad bolt=\quad 5-0.1=4.9\quad meters.$$
A boy standing on an open car throws a ball vertically upwards with a velocity of $$9.8\ m/s$$, while moving horizontally with uniform acceleration of $$1\ m/s^{2}$$ starting from rest. The ball will fall behind the boy on the car at a distance of
Report Question
0%
$$1\ m$$
0%
$$2\ m$$
0%
$$3\ m$$
0%
$$4\ m$$
Explanation
Initial velocity of the ball
$$u=9.8 m/s$$
$$v=0$$ (at the top)
So, $$v= u-gt$$
$$0= 9.8 - 9.8 t$$
$$t=1$$
So, to reach the height point the ball takes $$1$$ sec.
So, time taken by the ball to come back
$$=2t = 2sec$$
in $$2$$ sec, the car advances
$$S=ut+ \frac{1}{2}+t^2$$
$$= 0 + \frac{1}{2} 1\times4 = 2m$$
So the ball will fall $$2$$ m behind the car.
A parachutist after bailing out falls for $$10$$ s without friction. When the parachute opens he descends with an acceleration of $$2\ m/s^{2}$$ against his direction and reached the ground with $$4$$ $$m/s$$. From what height he has dropped himself? $$(g=10\ m/s^{2})$$
Report Question
0%
$$500$$ m
0%
$$2496$$ m
0%
$$2996$$ m
0%
$$4296$$ m
Explanation
$$u=gt\Rightarrow u=10g$$
$$h_{1}=\dfrac{1}{2}gt^{2}$$
$$h_{2}=\dfrac{v^{2}-u^{2}}{-2a}$$
total height $$=h_{1}+h_{2}$$
$$v = gt_{1} (100)^{2} = u^{2} +2 \times 10 S_{1}$$
$$v = 10 \times 10 = 100 m/s. [S_{1} = 500 mt]$$
$$4 = 100-2 t_{2}$$
$$t_{2} = 48 sec$$
$$\Rightarrow S_{2} = 100 t_{2} - 1/2 \times 2 t_{2}^{2}$$
$$ = 4800-(48)^{2}$$
$$ = 2496 $$ mt.
$$S_{1} + S_{2}= 2996$$ mt
A $$5\ kg$$ stone falls from a height of $$1000\ m$$ and penetrates $$2\ m$$ in a layer of sand. The time of penetration is
Report Question
0%
14.285 s
0%
0.0285 s
0%
7.146 s
0%
0.285 s
Explanation
Speed of the stone just before touching the sand is
$$v=\sqrt {2gh}=\sqrt {2\times 9.81\times 1000}=140.07m/s$$
Deceleration of the stone due to sand is
$$v^2-u^2=-2as$$
$$0-140.07^2=-2(a)(2)$$
$$a=4905m/s^2$$
We have initial speed and the deceleration, from
$$v=u-at$$
$$u=at$$
$$\displaystyle t=\frac {u}{a}=\frac {140.07}{4905}=0.0285s$$
Option B.
A ball of mass $$100$$ gm is projected vertically upward from the ground with a velocity of $$50 ms^{-1}$$ . At the same time another identical ball is dropped from a height of $$100$$ m to fall freely along the same path as that followed by the first ball. After some time the two balls collide, stick together and finally fall to the ground. The time taken by the combined mass to fall to the ground is approximately $$(g=10 ms^{-2})$$
Report Question
0%
4.5 s
0%
6.5 s
0%
9 s
0%
13 s
Explanation
$${V_{1}}^{2} = (50)^{2} - 2gx $$
$${V_{2}}^{2} = 2g(100 - x)$$
$$\Rightarrow {V_{2}}^{2} = 2g(100) + {V_{1}}^{2} - (50)^{2}$$
$$\Rightarrow {V_{2}}^{2} - {V_{1}}^{2} = 2000 - 2500$$
$$\Rightarrow (V_{1} + V_{2})(V_{1} - V_{2}) = \sqrt{-500}$$
$$V_{1} = 50 - gt$$
$$V_{2} = gt$$
$$\Rightarrow V_{1} + V_{2} = 50$$
$$\Rightarrow V_{1} - V_{2} = +10$$
$$V_{1} = 30 m/s (\uparrow)$$
$$V_{2} = 230 m/s (\downarrow)$$
$$\Rightarrow x =\dfrac{900-2500}{20} = 80$$ m
$$80 = -5t + \dfrac{1}{2}gt^{2}$$
$$t^{2} - t - 16 = 0$$
$$t = \dfrac{+1\pm \sqrt{1 + 64}}{2} = \dfrac{1\pm \sqrt{65}}{2}$$
$$=\dfrac{\sqrt{65}+1}{2}\approx 4.5$$ sec
A ball is projected from the bottom of a tower and is found to go above the tower and is caught by the thrower at the bottom of the tower after a time interval $$t_{1}$$. An observer at the top of the tower finds the same ball go up above him and then come back to his level in a time interval $$t_{2}$$ .The height of the tower is
Report Question
0%
$$\frac{1}{2}gt_{1}t_{2}$$
0%
$$\dfrac{gt_{1}t_{2}}{8}$$
0%
$$\dfrac{g}{8}\left ( t^{2}_{1}-t^{2}_{2} \right )$$
0%
$$\dfrac{g}{2}(t_{1}-t_{2})^{2}$$
Explanation
$$v=u+at\\ \Rightarrow u=\dfrac { g{ t }_{ 1 } }{ 2 } \\ s=ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 }\\ \Rightarrow s=\dfrac { g{ t }_{ 1 } }{ 2 } \dfrac { ({ t }_{ 1 }-{ t }_{ 2 }) }{ 2 } -\dfrac { g }{ 2 } \dfrac { { ({ t }_{ 1 }-{ t }_{ 2 }) }^{ 2 } }{ 4 } \\ \Rightarrow s=\dfrac { g({ t }_{ 1 }^{ 2 }-{ t }_{ 2 }^{ 2 }) }{ 8 } $$
A particle starts from rest with uniform acceleration $$a$$. Its velocity after $$n$$ second is $$v$$. The displacement of the body in the last two second is
Report Question
0%
$$\dfrac{2v\left ( n-1 \right )}{n}$$
0%
$$\dfrac{v\left ( n-1 \right )}{n}$$
0%
$$\dfrac{v\left ( n+1 \right )}{n}$$
0%
$$\dfrac{2v\left ( 2n+1 \right )}{n}$$
Explanation
We know,
At $$t = n $$sec
$$s_n = \dfrac{1}{2}an^2$$
$$v^2 = 2as_n$$
$$= 2a\cdot \dfrac{1}{2}an^2$$
$$= a^2n^2$$
$$\therefore v = an $$
$$\therefore a = \dfrac{v}{n}$$ ...(i)
Now, $$s_n-s_{n-2} = \dfrac{1}{2}a\left[n^2-(n-2)^2\right]$$
$$= \dfrac{1}{2}a\left[4n-4\right]$$
$$= 2a(n-1)$$ from (i)
$$s_n - s_{(n-2)} = \displaystyle \dfrac{2v(n-1)}{n}$$
A ball is dropped from a high rise platform at $$t =0$$ starting from rest. After $$6$$ seconds another ball is thrown downwards from the same platform with a speed $$v$$. The two balls meet at $$t = 18\ s$$. What is the value of $$v$$ ?
(take $$g = 10\ m/s^{2}$$)
Report Question
0%
$$75\ m/s$$
0%
$$55\ m/s$$
0%
$$40\ m/s$$
0%
$$60\ m/s$$
Explanation
Given , At t= 18 s ball meets
$$\Rightarrow $$ Distance covered by ball 1$$(S_1)$$ =
Distance covered by ball 2
$$(S_2)$$
Now, Using $$ S= ut +\dfrac{1}{2}at^2$$
$$ S_1 = \dfrac{1}{2}gt^2 = \dfrac{1}{2} 10(18)^2 = 1620m$$
And, $$S_2 = vt+\dfrac{1}{2}gt^2 = 12v + 720$$
But, $$ S_1 = S_2 $$
$$\Rightarrow 1620 = 720 +12v$$
$$\Rightarrow 900 = 12v$$
$$\Rightarrow v= 75 m/s$$
Therefore, A is correct option.
Two balls $$A$$ and $$B$$ are thrown with same velocity $$u$$ from the top of a tower Ball $$A$$ is thrown
vertically upward and the ball $$B$$ is thrown vertically downward. Choose the correct statement.
Report Question
0%
Ball $$B$$ reaches the ground with greater velocity
0%
Ball$$A$$ reaches the ground with greater velocity
0%
Both the balls reach the ground with same velocity
0%
Cannot be interpreted
Explanation
We know,
The direction of motion while touching the ground will be vertically downwards,
Now coming to the speed,
Speed is nothing but a reflection of Kinetic energy of a body,
More the kinetic energy more the speed.
Taking the ball-earth system,
Mechanical energy is conserved,
Gain in Kinetic energy=loss in potential energy
And as both the balls are at same height,
Irrespective of the initial direction of throw,
Both will lose same potential energy and gain the same Kinetic energy,
Hence having the same velocity when they are near the ground.
Option $$\textbf C$$ is the correct answer
A stone is dropped into a well in which the water level is h below, the top of the well. If $$v$$ is velocity of sound, then time $$T$$ after which the splash is heard is equal to
Report Question
0%
$$\dfrac{2h}{v}$$
0%
$$\sqrt{\dfrac{2h}{v}}+\dfrac{h}{g}$$
0%
$$\sqrt{\dfrac{2h}{g}}+\dfrac{h}{v}$$
0%
$$\sqrt{\dfrac{h}{2g}}+\dfrac{2h}{v}$$
Explanation
Time taken for the sound to reach to the top = Time taken for the stone to reach the ground + Time taken for the sound to reach the top.
$$h = ut + \dfrac{1}{2}at^2 $$ (u is zero since the stone is freely falling)
$$h = \dfrac{1}{2}at^2 $$
$$ t_g = \sqrt{\dfrac{2h}{g}}$$
Time taken for the sound to reach the top is
$$t_c = \dfrac{h}{v} $$
Total time taken $$= \sqrt{\dfrac{2h}{g}} + \dfrac{h}{v} $$
A truck running along a straight line increases its speed uniformly from $$30$$m/s to $$60$$m/s over a time interval $$1$$ min. The distance travelled during this time interval is
Report Question
0%
$$900$$ m
0%
$$1800$$ m
0%
$$2700$$ m
0%
$$3600$$ m
Explanation
Acceleration of the truck $$= \dfrac{v_2 - v_1}{t_2-t_1} $$
a $$= \dfrac{60 - 30}{60} $$
$$= 0.5\ m/s^2 $$
Distance travelled during this time interval is
s $$= ut + \dfrac{1}{2}at^2 $$
$$ = 30 \times 60 + \dfrac{0.5 \times 60^2}{2}$$
$$= 2700\ m $$
If a body is thrown vertically upward and rises to a height of 10 m, then time taken by the body to reach the highest point is
Report Question
0%
1.043 s
0%
1.43 s
0%
1.024 s
0%
none of these
Explanation
Using the equation of motions
$$v^2 = u^2 - 2gh $$
At maximum height , v = 0
$$u^2 = 2gH $$
$$u^2 = 2 \times 9.8 \times 10 $$
$$u^2 = 196 $$
$$u = 14 m/s $$
$$v = u - gt $$
$$t = \frac{u}{g} $$
$$t = \frac{14}{9.8} $$
$$ = 1.43 s$$
A particle is thrown upwards from ground. It experiences a constant air resistance force which can produce a retardation of $$2 m/s^{2}.$$ The ratio of time of ascent to the time of descent is $$:\left[ g = 10\ m/s^{2} \right] $$
Report Question
0%
$$1 : 1$$
0%
$$\sqrt{\dfrac{2}{3}}$$
0%
$${\dfrac{2}{3}}$$
0%
$$\sqrt{\dfrac{3}{2}}$$
Explanation
Let a be the retardation produced by resistive force, $$t_{a}\ and\ t_{d}$$ be the time ascent and descent respectively.
Let the particle rise upto a height $$h$$ and projection velocity be $$u$$.
For upward motion,
$$0 = u - (a+g)t_a ..........(i)$$
$$0 = u^2 -2(a+g)h..........(ii)$$
For downward motion,
$$v_2 = 0 + (g-a)t_d.........(iii)$$
$$v_2^2 = 0 + 2(g-a)h.......(iv)$$
From $$(ii)\ \&\ (iv),$$
$$v_2 = u_2 \sqrt{\cfrac{ g-a}{g+a}}...........(v)$$
From $$(i)\ \&\ (iii),$$
$$\cfrac{t_a}{t_d}= \cfrac{u(g-a)}{v_2(g+a)}............(vi)$$
From $$(v)\ \&\ (vi),$$
$$\cfrac{t_a}{t_d} = \sqrt{\cfrac{g-a}{g+a}}$$
$$=\sqrt { \cfrac{2}{3}}$$
A car starts from rest and moves along the x-axis with a constant acceleration $$5\ ms^{-2}$$ for $$8\ s$$. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from rest?
Report Question
0%
160 m
0%
240 m
0%
320 m
0%
400 m
Explanation
The distance travelled in first eight seconds $$=x_{1}=0+\frac{1}{2}*5*8^{2}=160\ m$$
At this point the velocity $$ v=u+at=0+ 5\times 8 =40\ ms^{-1}$$
Therefore, the distance covered in last four seconds, $$x_{2}= 40 \times 4=160 m$$
Thus, the total distance $$x=x_{1}+x_{2}= 160+160 =320 m$$
An object starting from rest travels 20 m in first 2 s and 160 m in next 4 s.
What will be the velocity( in $$m/s$$) after 7 s from the start.
Report Question
0%
0
0%
10
0%
65
0%
70
Explanation
From $$s=ut+\dfrac{1}{2}at^2$$
In first 2 s,
$$s=20 m$$
$$20 =\frac{1}{2}a(2)^2=2 a$$
$$a = 10\ m/s^2$$
Velocity at the end of 2 s is
$$v = u+at = 0 + 10 \times 2 = 20 m/s$$
In next 4s, $$s=ut +\dfrac{1}{2} a't^2$$
$$160 = 20 \times 4+\frac{1}{2}a'(4)^2$$
$$80 = 8 a'$$
$$a' = 10\ m/s^2$$
It shows that acceleration is uniform. From $$v=u + at$$
$$ v=0+10 \times 7 = 70\ m/s$$
Answer: (D) 70
A particle moves in a straight line with a constant acceleration. It changes its velocity from $$ 10 ms^{-1} \ to \ 20 ms^{-1}$$ while passing through a distance $$135m$$ in $$t$$ second. The value of t is
Report Question
0%
$$12$$
0%
$$9$$
0%
$$10$$
0%
$$1.8$$
Explanation
$$u=10m/s,v=20m/s,x=130m$$
$$a=\cfrac{v-u}{t}=\cfrac{20-10}{t}=\cfrac{10}{t}$$
$$s=ut+\cfrac{1}{2}at^2\\\implies 135 =10t+\cfrac{1}{2}\cfrac{10}{t}t^2\\\implies t=9s$$
A particle is projected vertically upwards with a speed $$u$$.
Report Question
0%
When it rises to half its maximum height its velocity is $$\dfrac{u}{2}$$.
0%
Time taken to rise to half its maximum height is half the time taken to reach maximum height.
0%
Time taken to rise to three fourth its maximum height is half the time taken to reach its maximum height.
0%
The acceleration of the particle when it reaches its maximum height is zero.
Explanation
If $$ h$$ is the maximum height then $${ h= }\dfrac { { u }^{ 2 } }{ 2g } $$
Time taken to reach height $$ h$$, $$ t=\dfrac { u }{ g } $$
Let $$v$$ be the speed at $$ \dfrac{3h}{4}$$ then $$ { v }^{ 2 }={ u }^{ 2 }-2g\left( \dfrac { 3h }{ 4 } \right) \Rightarrow v=\dfrac { u }{ 2 } $$
let $$ t'$$ be the time taken to reach $$ \dfrac{3h}{4}$$ then $$ \dfrac { u }{ 2 } =u-gt'\Rightarrow t'=\dfrac { u }{ 2g } =\dfrac { t }{ 2 } $$
A stone is dropped from a certain height, which can reach the ground in 5 s. After 3 s of its fall, it is stopped and again allowed to fall. Then, the time taken by the stone to reach the ground for the remaining distance is :
Report Question
0%
3 s
0%
4 s
0%
2 s
0%
None of these
Explanation
$$T_d = 5s, H= \displaystyle \dfrac{1}{2} \times g \times 25 = 12.5 g$$
$$t_1 = 3s ; h = \displaystyle \dfrac{1}{2} \times g \times 9 =4.5 g$$
$$(H-h)=8g = \displaystyle \dfrac{1}{2} \times g \times t^2 \Rightarrow t = \sqrt{16} = 4 s$$
A body falls from a height of 200 m. If gravitational attraction ceases after 2 s, further time taken by it to reach the ground is $$(g = 10 m s^{-2})$$.
Report Question
0%
5 s
0%
9 s
0%
13 s
0%
17 s
Explanation
$$t= 2s$$
$$h_1 = \displaystyle \dfrac{1}{2} gt^2$$
$$h_1 \displaystyle = \dfrac{1}{2} \times 10 \times 4 = 20 m$$
After falling through 20 m, g becomes zero remaining distance to be travelled
(d$$ = 200 - 20 = 180 m$$)
and
$$v = gt = 10 \times 2 = 20 m s^{-1}$$
$$\therefore t \displaystyle = \dfrac{d}{v} = \dfrac{180}{20} = 9s$$
A body thrown vertically up with a velocity 'u' reaches the maximum height 'h' after 'T' second. The correct statement among the following is :
Report Question
0%
At a height $$\displaystyle \dfrac{h}{2}$$ from the ground its velocity is $$\displaystyle \dfrac{u}{2}$$
0%
At a time T its velocity is 'u'
0%
At a time '2T' its velocity is -u
0%
At a time 2T its velocity is -6u
Explanation
Time taken by the body to reach the maximum height is given as $$T$$ and the velocity of the body at the time of throw is $$u$$ (upwards). After further time $$T$$, the body reaches to the ground having same velocity $$u$$ in the downward direction as that of initial velocity at the time of throw. Thus, at the instant $$2T$$, the velocity of the body is $$-u$$ (where minus sign represents the velocity to be in downward direction).
Two stones are dropped down simultaneously from different heights. At the starting time, the distance between them is 30 cm. After 1 s, the distance between the two stones will be $$(g = 10 m s^{-2}).$$
Report Question
0%
$$10 cm$$
0%
$$20 cm$$
0%
$$30 cm$$
0%
$$0 cm$$
Explanation
After 2 s or any difference in seconds, separation will be 30 cm only as both bodies covers same distance for same time interval under gravity.
A freely falling body travels with uniform acceleration .
Report Question
0%
True
0%
False
Explanation
When a body is in free fall motion, acceleration due to gravity $$(g = 10m/s^2)$$ always acts in the downward direction. So, the given statement is true that a freely falling body travels with uniform acceleration.
A gun is fired at a target. At the moment of firing, the target is released and allowed to fall freely under gravity. Then the bullet :
(Assume zero air resistance)
Report Question
0%
Misses the target by passing above it
0%
Hits the target
0%
Misses the target by passing below it
0%
May or may not hit
Explanation
Initial vertical component of the velocities (in downward direction) of both the bullet and the target are zero. So, the bullet and the target fall down by equal amount and thus the bullet hits the target.
A person, seated in a train under motion, is at rest with reference to :
Report Question
0%
The train.
0%
A person watching him from the front seat.
0%
A car moving in the same direction and speed of the train.
0%
Trees on ground.
Explanation
The person, seated in a train under motion, will be at rest with reference to the object whose relative velocity is zero. So, that person is at rest with respect to the train itself, another person watching him from the front seat and the car moving in the same direction with same speed.
A body A is thrown vertically upwards with such a velocity that it reaches a maximum height of $$h$$ in time $$t$$. Simultaneously another body B is dropped from height $$h$$. It strikes the ground and does not rebound. The velocity of A relative to B v/s time graph is best represented by (upward direction is positive)
Report Question
0%
0%
0%
0%
Explanation
For the first half, when ball A goes up
$$V_A=U_A-gT$$
$$V_B=-gT$$
$$V_{AB}=V_A-V_B$$
$$V_{AB}=U$$.......$$(1)$$
For the second half, when ball A comes down
$$V_A=-gT$$
$$V_B=0$$
$$V_{AB}=V_A-V_B$$
$$V_{AB}=-gT$$........$$(2)$$
Hence from the equation $$(1)$$ we observe that relative velocity is independent of $$T$$. Which will be valid till ball A reaches maximum height point. It is given time taken in reaching $$h$$ is $$t$$.
Hence for the time $$t$$, $$V_{AB}$$ will be constant, whereas after time $$t$$,
relative velocity changes its direction and increases afterwards.
The ratio of time taken by two cars P, Q starting from rest moving along a straight road with equal accelerations is $$\sqrt{2}$$ : 1, then the :
Report Question
0%
Final velocity of car P > final velocity of car Q.
0%
Final velocity of car P < final velocity of car Q.
0%
Ratio of $$V_P$$ to $$V_Q$$ is 2 : $$\sqrt{2}$$.
0%
Ratio of distance travelled by car 'P' to car 'Q' is 2 : 1
Explanation
$$t_P : t_Q = \sqrt{2} : 1 ; u_P = u_Q =0$$
$$\displaystyle \dfrac{v_P}{v_Q} = \dfrac{t_P}{t_Q} = \dfrac{\sqrt{2}}{1} \Rightarrow v_P = \sqrt{2}V_Q$$
$$\therefore $$ Final velocity of car P > final velocity of Q
$$v_P : v_Q = 2 : \sqrt{2}$$
$$\displaystyle \dfrac{S_P}{S_Q} = \dfrac{t_P^2}{t_Q^2} = \dfrac{(\sqrt{2})^2}{(1)^2} = \dfrac{2}{1}$$
A body standing on a long railroad car throws a ball straight upwards, the car is moving on the horizontal road with an acceleration $$1\ m{ s }^{ -2 }$$. The vertical velocity given is $$9.8\ m{ s }^{ -1 }$$. How far behind the boy the ball will fall on the railroad car?
Report Question
0%
$$1\ m$$
0%
$$\cfrac {3}{2}\ m$$
0%
$$\cfrac {7}{4}\ m$$
0%
$$2\ m$$
Explanation
For vertical direction we see that the relative distance travelled is zero. Thus
$${ s }_{ rel }=ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 }\\ \Rightarrow 0=9.8t-\dfrac { 1 }{ 2 } { 9.8t }^{ 2 }\\ \Rightarrow t=2s.$$
Now for horizontal direction we have
$$s=ut+\dfrac { 1 }{ 2 } a{ t^2 }\\ s=0+\dfrac { 4 }{ 2 } =2m$$
The acceleration of a particle as seen from two frames $$S_{1}$$ and $$S_{2}$$ has equal magnitude $$5\ ms^{-2}$$.
Report Question
0%
The frames must be at rest with respect to each other.
0%
The frames may be moving with respect to each other but neither should be accelerated with respect to the other.
0%
The acceleration of frame $$S_{2}$$ with respect to $$S_{1}$$ be $$0$$ or $$10\ ms^{-2}$$
0%
The acceleration of frame $$S_{2}$$ with respect to $$S_{1}$$ lies between $$0$$ and $$10\ ms^{-2}$$
Explanation
Using parallelogram law,
Let,
$$\vec {S_1} \,\,\, \vec {S_2}$$ be the arms of the parallelogram,
Given,
$$\vec {S_1} = \vec {S_2}$$
the angle between these two vectors may lie between $$0^o$$ to $$180^o$$
Let us find the resultant taking $$\theta = 0^o \,\,\, and \,\,\,\, 180^o$$
$$R = \sqrt{{S_1}^2 + {S_2}^2 - 2.{S_1}{S_2}cos\theta}$$
$$\therefore R = \sqrt{5^2 + 5^2 - 2.(5)(5)cos{0^o}} = 0$$
AND
$$R = \sqrt{5^2 + 5^2 - 2.(5)(5)cos{180^o}} = 10$$
$$\therefore $$ The acceleration of frame $$S_{2}$$ with respect to $$S_{1}$$ lies between $$0$$ and $$10ms^{-2}$$
A flower pot falls off a window sill and falls past the window below. It takes $$0.5s$$ to pass through a $$2.0m$$ high window. Find how high is the window sill from the top of the window?
Report Question
0%
$$10cm$$
0%
$$7.5cm$$
0%
$$11.25cm$$
0%
$$15cm$$
Explanation
For motion from $$B$$ to $$C$$
$$t=0.5s, s=BC=2\ m$$
Velocity at $$B$$ is $$ { u }_{ B }$$
$$ s={ u }_{ B }t+\dfrac { 1 }{ 2 } g{ t }^{ 2 }\quad \Rightarrow \quad { u }_{ B }=1.5\ m/s$$
For motion from $$A$$ to $$B$$
$$ s=AB=h, { u }_{ A }=0\quad { v }_{ B }{ =u }_{ B }=1.5\ m/s$$
$$ { v }_{ B }^{ 2 }={ u }_{ A }^{ 2 }+2gh\quad $$
or
$$ \left( 1.5 \right)^2 =0+2\left( 10 \right) h\quad or\quad h=0.1125m=11.25cm$$
A body projected vertically upward with a velocity $$v$$ at $$t=0$$ is found at a height $$h$$ after $$1\ s$$ and a further $$6\ s$$, it is found at the same height $$\displaystyle (g=10\ ms^{-2})$$. Then
Report Question
0%
$$h$$ is $$30\ m$$
0%
$$v$$ is $$40\ ms^{-1}$$
0%
maximum height is
$$ 80\ m$$
0%
distance moved in
$$ 5th$$ second is $$5\ m$$
Explanation
Let that height be $$h$$.
$$Thus,\ h = v(1) - \dfrac { 1 }{ 2 } \times10\times{ (1) }^{ 2 }= v(7) - \dfrac { 1 }{ 2 }\times 10\times{ (7) }^{ 2 }\\ This\ gives\ v= 40\ m/s\quad and\ h =35\ m\\ Now,\quad max\quad height\quad means\quad velocity\quad becomes\quad 0\\ Hence,\quad { H }_{ max }= \dfrac { { 0 }^{ 2 } -{40}^{2}}{2\times (-10)} =80\quad m\\ 0 =40 -10\ t\quad gives\quad time\quad to\quad reach\quad this\quad height\quad as\quad 4\ seconds.\\ Thus\quad in\quad 5th\quad second\quad it\quad starts\quad from\quad zero\quad velocity.\\ Thus\quad distance, s = 0 +\dfrac { 1 }{ 2 }\times 10\times{ 1 }^{ 2 }=5\ m$$
A particle is projected vertically upward with a speed $$u$$.
Report Question
0%
When it rises to half its maximum height its velocity is $$\dfrac{u}{2}$$
0%
Time taken to rise to half its maximum height is half the time taken to reach maximum height
0%
Time taken to rise to three fourth its maximum height is half the time taken to reach maximum height
0%
The acceleration of the particle when it reaches its maximum height is zero.
Explanation
Time taken by particle to reach maximum height,
v = u - gt
0 = u- gt
t = u/g
we know,
H = $$\dfrac { { u }^{ 2 } }{ 2g } $$
$$\dfrac { 3H }{ 4 } $$ = $$\dfrac { 3{ u }^{ 2 } }{ 8g } $$
$$h = ut - \dfrac { 1 }{ 2 } g{ t }^{ 2 }$$
$$\dfrac { 3{ u }^{ 2 } }{ 8g } = ut -\dfrac { 1 }{ 2 } g{ t }^{ 2 }$$
$$g{ t }^{ 2 }$$-2ut+$$\dfrac { 3{ u }^{ 2 } }{ 4g }=0 $$
on solving quadratic, we get
$$t = \dfrac{u}{2g}$$
From the top of a tower, two stones whose masses are in the ratio $$1:2$$ are thrown, one straight up with an initial speed $$u$$ and the second straight down with same speed $$u$$. neglecting air resistance,
Report Question
0%
the heavier stone hits the ground with a higher speed
0%
the lighter stone hits the ground with a higher speed
0%
both the stones will have same speed when they hit the ground
0%
the speed cannot be determined with the given data
Explanation
For motion under gravity, speed is independent of masses
The net vertical displacement for both stone are same. Also they are launched with same initial speed $$u$$, therefore both the stone will have same speed when they hit the ground.
A ball is thrown up with certain velocity so that it reaches a height $$h$$. Find the ratio of the times in which it is at $$\dfrac{h}{3}$$.
Report Question
0%
$$\dfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } +1 } $$
0%
$$\dfrac { \sqrt { 3 } -\sqrt { 2 } }{ \sqrt { 3 } +\sqrt { 2 } } $$
0%
$$\dfrac { \sqrt { 3 } -1 }{ \sqrt { 3 } +1 } $$
0%
$$\cfrac {1} {3}$$
Explanation
At maximum height, velocity is zero so, $$ 0={ u }^{ 2 }-2gh\quad or\quad u=\sqrt { 2gh } $$
As $$ s=ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 }$$ so $$, \dfrac { h }{ 3 } =\sqrt { 2gh } t-\dfrac { 1 }{ 2 } g{ t }^{ 2 }$$
$$ g{ t }^{ 2 }-2\sqrt { 2gh } t+\dfrac { 2h }{ 3 } =0\quad or\quad t=\dfrac { 2\sqrt { 2gh } \pm \sqrt { 8gh-\dfrac { 8gh }{ 3 } } }{ 2g } $$
$$ \dfrac { { t }_{ 1 } }{ { t }_{ 2 } } =\dfrac { 2\sqrt { 2gh } -\sqrt { 8gh-\dfrac { 8gh }{ 3 } } }{ 2\sqrt { 2gh } +\sqrt { 8gh-\dfrac { 8gh }{ 3 } } } =\dfrac { \sqrt { 3 } -\sqrt { 3-1 } }{ \sqrt { 3 } +\sqrt { 3-1 } } =\dfrac { \sqrt { 3 } -\sqrt { 2 } }{ \sqrt { 3 } +\sqrt { 2 } } $$
Which of the following statements regarding motion of particle is true?
Report Question
0%
The motion between $$A$$ and $$B$$ is known
0%
The motion between $$A$$ and $$B$$ is erratic
0%
The motion between $$A$$ and $$B$$ may have been steady or erratic
0%
The motion between $$A$$ and $$B$$ is steady
Explanation
the motion between A and B is steady if the velocity of the particle is constant and erratic if it is variable. From the question it is not clear what happens with the particle during the motion so we cannot say whether the particle is in erratic or steady motion.
Engine of a vehicle can give it an acceleration of $$1\;m{ s }^{ -2 }$$ and its brakes can retard it at $$3\;m{ s }^{ -1 }$$. The minimum time in which the vehicle can make a journey between stations $$A$$ and $$B$$ having a distance of $$1200\;m$$ is
Report Question
0%
$$55.6\;s$$
0%
$$65.6\;s$$
0%
$$50.6\;s$$
0%
$$56.5\;s$$
Explanation
Let $$x$$ be the distance traveled until maximum velocity (V) and $$y$$ be the remaining distance
we can write
$$ { v }^{ 2 }-0=2.1.x\quad \Rightarrow \quad x=\dfrac { { v }^{ 2 } }{ 2 }$$
Similarly
$$0-{ v }^{ 2 }=-6y\quad \Rightarrow \quad y=\dfrac { { v }^{ 2 } }{ 6 } \\ x+y=1200=\dfrac { { 2v }^{ 2 } }{ 3 } \Rightarrow { v }=30\sqrt { 2 }$$
Now, using equation $$ v=u+at$$
we get
$${ t }_{ 1 }=30\sqrt { 2 } \quad \& \quad { t }_{ 2 }=10\sqrt { 2 } \\ { t }_{ 1 }+{ t }_{ 2 }=40\sqrt { 2 } =56.5$$ s
An object may have
(I) varying speed without having varying velocity.
(II) varying velocity without having varying speed.
(III) non-zero acceleration without having varying velocity.
(IV) non-zero acceleration without having varying speed.
Report Question
0%
I and II are correct.
0%
II and III are correct.
0%
II and IV are correct.
0%
None of the above.
Explanation
Speed is a scalar quantity while velocity is a vector quantity.
An object can have constant speed but velocity will change since velocity is a vector quantity.
Without having a varying speed non-zero acceleration. Uniform circular motion is an example of options B and D
Thus, B and D options are correct
A body travels $$200\ cm$$ in the first two seconds and $$220\ cm$$ in the next four seconds. What will be the velocity at the end of $$7^{th}$$ second from the start?
Report Question
0%
$$10\ cm s^{ -1 }$$
0%
$$20\ cm{ s }^{ -1 }$$
0%
$$15\ cm{ s }^{ -1 }$$
0%
$$5\ cm{ s }^{ -1 }$$
Explanation
$$ s= ut + \dfrac { 1 }{ 2 } a t^2$$
For $$case\ 1.$$
$$ 200 = 2u + 2a$$
$$ u + a = 100 ....(1)$$
For $$case\ 2.$$ .i.e. next 4 seconds, we have distance $$= 420\ cm$$ and time $$= 6\ s$$
$$ 420 =6 u+18a$$
$$ u + 3a= 70 ....(2) $$
Solving equation $$1$$ and $$2,$$ we get,
$$u = 115\ cm/s$$ and $$a = -15\ cm/s^2$$
Now by
$$ v=u+ at$$
$$ v = 115 -15\times 7$$
$$ v =10\ cm/s $$
Hence answer is A.
A player throws a ball upwards with an initial speed of $$29.4\ m{ s }^{ -1 }$$. The height to which the ball rises and the time taken to reach the player's hands are
Report Question
0%
$$22.05\ m,38\ s$$
0%
$$44.1\ m,6\ s$$
0%
$$29.4\ m,6\ s$$
0%
$$54.5\ m,9\ s$$
Explanation
Maximum height $$ h=\dfrac { { u }^{ 2 } }{ g } =\dfrac { { \left( 29.4 \right) }^{ 2 } }{ 2\left( 9.8 \right) } =44.1m$$
Time taken $$ t=2\sqrt { \dfrac { 2h }{ g } } =2\sqrt { \dfrac { 2\left( 44.1 \right) }{ 9.8 } } =6s$$
A particle experiences a fixed acceleration for $$6$$ seconds after starting from rest. It cover a distance of $${ s }_{ 1 }$$ in first two seconds, $${ s }_{ 2 }$$ in the next 2 seconds and $${ s }_{ 3 }$$ in the last 2 seconds then $${ s }_{ 3 }:{ s }_{ 2 }:{ s }_{ 1 }$$ is
Report Question
0%
$$1:3:5$$
0%
$$5:3:1$$
0%
$$1:2:3$$
0%
$$4:3:2:1$$
Explanation
Using the equations
$$s=ut+\frac { 1 }{ 2 } a{ t }^{ 2 }\\ v=u+at$$
We get
For $$t=0$$ to $$t=2$$
$$u=0\quad m/s\\ s=2a\quad m$$
For $$t=2$$ to $$t=4$$
$$u=2a\quad m/s\\ s=6a\quad m$$
For $$t=4$$ to $$t=6$$
$$u=4a\quad m/s\\ s=10a\quad m$$
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 11 Engineering Physics Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
