MCQ Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 1

A lizard, at an initial distance of

$21$ cm behind an insect, moves from rest with an
acceleration of

$2 cm/s^{2}$ and pursues the insect which is crawling uniformly along a
straight line at a speed of

$20$ cm/s. Then the lizard will catch the insect after

0%
$20$ s
0%
$1$ s
0%
$21$ s
0%
$24$ s

Explanation

Let t be the time in which the lizard will catch the insect.

Distance traveled by lizard $=$ 21 + distance traveled by insect.

$ut+\frac { 1 }{ 2 } a{ t }^{ 2 }=21+vt$

$0+\frac { 1 }{ 2 } 2{ t }^{ 2 }=21+20t$

${ t }^{ 2 }-20t-21=0$

solving this $t= 21 , -1$

t cannot be -ve.

$t=21 sec.$

Two points

$A$ and

$B$ move from rest along a straight line with same acceleration

$f$ and

$f'$ respectively. If

$A$ takes

$m\ \text{sec}$ more than

$B$ and describes

$n$ units more than

$B$ in acquiring the same speed then

0%
$(\mathrm{f}-\mathrm{f}')\mathrm{m}^{2}=\mathrm{f}\mathrm{f}'\mathrm{n}$
0%
$(\mathrm{f}+\mathrm{f}')\mathrm{m}^{2}=\mathrm{f}\mathrm{f}'\mathrm{n}$
0%
$\displaystyle \frac{1}{2}(\mathrm{f}+\mathrm{f}')\mathrm{m}=\mathrm{f}\mathrm{f}'\mathrm{n}^{2}$
0%
$(f^{'}-f)n=\frac{1}{2}ff^{'}m^{2}$

Explanation

$\mathrm{v}^{2}=2\mathrm{f}(\mathrm{d}+\mathrm{n})=2\mathrm{f}'\mathrm{d}$

$\mathrm{v}=\mathrm{f}'(\mathrm{t})=(\mathrm{m}+\mathrm{t})\mathrm{f}$

eliminate $\mathrm{d}$ and $\mathrm{m}$ we get $(f^{'}-f)n=\frac{1}{2}ff^{'}m^{2}$

Two stones are thrown up simultaneously from the edge of a cliff

$240\ m$ high with initial speed

$10\ m/s$ and

$40\ m/s$ respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first ?
(Assume stones do not rebound after hitting the ground and neglect air resistance, take

$g = 10\ m /s^2$ )
(The figures are schematic and not drawn to scale.)

Explanation

For the first ball,

$-240 = 10t -\frac{1}{2}gt^2$

$\therefore 5t^2-10t -240$

$\therefore t^2-2t-48 =0$

$\Rightarrow t=8,-6$
The 1st particle will reach the ground in 8 secs.
Upto 8 secs, the relative velocity between the particles is 30m/sec and the relative acceleration is zero..
For the 2nd particle,

$-240 = 40t -5t^2$

$\Rightarrow t^2 -8t -48 =0$

$t =12\: secs$
The second particle will strike the ground in 12 secs.
Option(A): Linear behaviour after 8 secs, hence can be eliminated.
Option(B) shows increase until 12 secs , hence is not correct. after 8 secs, separation will decrease.
Option(c) is the correct one.

STATEMENT-1: For an observer looking out through the window of a fast moving train, the nearby objects appear to move in the opposite direction to the train, while the distant objects appear to be stationary.

STATEMENT-2: If the observer and the object are moving at velocities

$\vec{\mathrm{V}}_{1}$ and

$\vec{\mathrm{V}}_{2}$ respectively with reference to a laboratory frame, the velocity of the object with respect to the observer is

$\vec{\mathrm{V}}_{2}-\vec{\mathrm{V}}_{1}$ .

0%
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
0%
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
0%
STATEMENT -1 is True, STATEMENT-2 is False
0%
STATEMENT -1 is False, STATEMENT-2 is True

Explanation

Both the statements are true,but correct explanation of statement one is

we,know that

${\theta}=\frac{arc}{radius}$

so value of

${\theta}$ will be more for nearby object and radius will be less wile value of

${\theta}$ will be less for distant object and radius will be more.

that is the reason behind this phenomenon and near one will move faster and distant object will move relatively less(i.e both are moving w.r.t any frame of refrence)

A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field

$\overrightarrow{E}$ . Due to the force

$q\overrightarrow{E}$ , its velocity increases from

$0$ to

$6\ m/s$ in one-second duration. At that instant, the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between

$0$ to

$3$ seconds are respectively:

0%
$1$ m/s, $3.5$ m/s
0%
$2$ m/s, $4$ m/s
0%
$1.5$ m/s, $3$ m/s
0%
$1$ m/s, $3$ m/s

Explanation

Acceleration toward +ve

x-direction in first case is a

$v-u=at$

$a=\frac{6}{1}=6m/s^{2}$

when field reserved acceleration changed to

$a=-6m/s^{2}$

$x=\frac{1}{2}at_{1}^{2}=\frac{1}{2}*6*1=3m$

$y=vt+\frac{1}{2}at_{2}^{2}=6*2+\frac{1}{2}(-6)*2^{2}$

$y=12-12=0$

$x_{2}=$ distance traveled is 1 sec

$x_{2}=6*1-\frac{1}{2}(6)*1$

$=3m$

total distance cover

$=x_{1}+2x_{2}=9m$

Total dispalcement

$=3m$

Average velocity =

$\dfrac{displacement \ cover}{time \ taken}=\dfrac{3}{3}=1 \ m/s$

Average speed

$=\dfrac{distance \ cover}{time \ taken}=\dfrac{9}{3}=3 \ m/s$

1007282_885693_ans_9a3ba2e9741b4a7e9de0341d79bc6978.png

A ball is dropped from a bridge

$122.5 m$ high. After the first ball has fallen for

$2 seconds$ , a second ball is thrown straight down after it, what must be the initial velocity of the second ball be, so that both the balls hit the surface of water at the same time?

0%
$26.1 {m}/{s}$
0%
$9.8 {m}/{s}$
0%
$55.5 {m}/{s}$
0%
$49 {m}/{s}$

Explanation

Time taken by the first object to reach the ground $= t$ , so

$122.5 = ut + \displaystyle\frac{1}{2} g{t}^{2}$

$122.5 = \displaystyle\frac{1}{2} \times 10 \times {t}^{2}$

$\Rightarrow t = 5 sec$ (approx)

Time to be taken by the second ball to reach the ground $= 5 2 = 3 sec$ .

If $u$ be its initial velocity then,

$122.5 = u \times 3 + \displaystyle\frac{1}{2} g{t}^{2} = 3u + \displaystyle\frac{1}{2} \times 10 \times 9$

$3u = 122.5 -45 = 77.5$

$u = 26$ (approx)

If a car at rest accelerated uniformly to a speed of

$144 {km}/{hour}$ in

$20 second$ it covers a distance:

0%
$400 m$
0%
$1440 m$
0%
$2880 m$
0%
$25 m$

Explanation

$u = 0, v = 144 {km}/{hour}= 144 \times \displaystyle\frac{5}{18} {m}/{sec} = 40 {m}/{sec}$

$v = u + at$

$\Rightarrow a = \displaystyle\frac{v-u}{t} = \displaystyle\frac{40-0}{20} = 2 {m}/{{sec}^{2}}$

$\therefore s = ut + \displaystyle\frac{1}{2} a{t}^{2}$

$= \displaystyle\frac{1}{2} \times 2 \times {\left(20\right)}^{2} = 400 m$

A ball is thrown vertically upwards with a given velocity such that it rises for T seconds (T > 1). What is the distance traversed by the ball during the last one second of ascent (in meters) ? (Acceleration due to gravity is g

$m/s^{2}$ ) .

0%
$v+g\dfrac{(1-2T)}{2}$
0%
$vT + \frac {1}{2} g [T^2 - (T-1)^2]$
0%
$\dfrac {g}{2}$
0%
$\frac {1}{2} g [T^2 - (T-1)^2]$

The accelerated motion of a body can occur :

0%
Due to change in its speed only.
0%
Due to change in direction of motion only.
0%
Due to change in both speed and direction of motion.
0%
Due to constancy of velocity.

Unit for the rate of change of velocity is

0%
m / s
0%
$m / s^2$
0%
N s
0%
N / s

Explanation

Velocity units

$= m/s$
Acceleration is the rate of change of velocity
a

$= \dfrac{change \ in \ velocity }{time }$
Therefore SI units of acceleration is

$ms^{-1}/s = \dfrac{m}{s^2}$

A body starts from rest and acquires a velocity

$10m s^{-1}$ in 2 s. Find the acceleration.

0%
$10 m s^{-2}$
0%
$20 m s^{-2}$
0%
$5 m s^{-2}$
0%
$40 m s^{-2}$

Explanation

Acceleration = Change in velocity per unit time.

Therefore, Acceleration

$a = \dfrac{10 m/s}{2 s} = 5 m/s^2$

A car starting from rest acquires a velocity

$180\ m s^{-1}$ in 0.05 h. Find the acceleration.

0%
$3600 m s ^{-2}$
0%
$9 m s ^{-2}$
0%
$2 m s ^{-2}$
0%
$1 m s ^{-2}$

Explanation

Acceleration,

$a = \dfrac{180 m/s}{0.05\times 3600 s} = 1 m/s^2$

Unit of acceleration is

0%
$m/s$
0%
$ms$
0%
$m/s^2$
0%
none of these

Explanation

acceleration

$= \dfrac{dv}{dt}$
Units of 'v' is m/s and 't' is seconds
Therefore units of acceleration is

$\dfrac{m}{s^2}$

A body initially moving with a velocity of

$5\ ms^{-1}$ attains a velocity of

$25\ ms^{-1}$ in

$5\ s$ , find the acceleration of body in

$ms^{-2}.$

0%
$8ms^{-2}$
0%
$7ms^{-2}$
0%
$4ms^{-2}$
0%
$3ms^{-2}$

Explanation

Acceleration

$= \dfrac{\text{Final velocity - initial velocity}}{\text{time taken}}$

Final velocity

$v = 25 m/s$
Initial velocity

$u = 5 m/s$
a

$= \dfrac{25 - 5}{5} m/s^2$
a

$= 4 m/s^2$

The velocity of a body can change

0%
if its acceleration is zero.
0%
if its acceleration is non-zero.
0%
Both A & B
0%
None of the above.

Which of the following is/are true about acceleration

0%
Acceleration is defined as the rate of change of velocity.
0%
Its SI unit is $m/s^2$ .
0%
Negative acceleration is called retardation
0%
All the above.

Explanation

Acceleration =

$\dfrac{Final\ velocity\ - Initial\ velocity }{time}$

Its SI unit is =

$ms^{-1}/s=m/s^2$

A negative acceleration acts opposite to the direction of motion of object and it reduces the velocity , it called retardation.

A block is released from rest at the top of a friction-less inclined plane 16 m long. It reaches the bottom 4 sec later. A second block is projected up the plane from the bottom. At the instant at which the block is released in such a way that it returns to the bottom simultaneously with the first block. The acceleration of each block on the incline is :

0%
$1 m/s^{2}$
0%
$2 m/s^{2}$
0%
$4 m/s^{2}$
0%
$9.8 m/s^{2}$

Explanation

Acceleration of both the blocks will be the same on the inclined plane

Let it be '

$a$ '

So, according the the equation of motion

$\frac { 1 }{ 2 } \times a\times { 4 }^{ 2 }=16\Rightarrow a=2m/{ s }^{ 2 }$

If the velocity of a body does not change, its acceleration is

0%
Zero
0%
Infinite
0%
Unity
0%
None of these

An object is thrown vertically upwards and rises to a height of 10 m. Calculate the velocity with which the object was thrown upwards :

0%
7 m/s
0%
14 m/s
0%
28 m/s
0%
56 m/s

Explanation

$v^2=u^2+2as$

$0=u^2+2(-9.8)(10)$

$u= \sqrt{2\times9.8\times10}= 14 m/s$

What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?

0%
Object is moving with uniform acceleration
0%
Object is moving with uniform speed
0%
Object is moving at rest
0%
None

Explanation

The correct answer is option (B).

$\bullet$

From the given v - t graph, it can be inferred that the object is

$\bullet$ If the speed-time graph is a straight line parallel to the time axis it means that on changing time speed remains constant.

The correct answer is an option (B).

When a ball is thrown vertically upwards, it reaches a maximum height of 5m. The initial velocity of the ball was ?

0%
5 m/s
0%
50 m/s
0%
10 m/s
0%
25 m/s

Explanation

Answer is C.

The equation is given as

${ v }^{ 2 }-{ u }^{ 2 }=2as$ .
where,
v = final velocity,
u = initial velocity,
a = acceleration due to gravity,

$10\quad m/{ s }^{ 2 }$ ,
s = h = height traveled.
Therefore, we can find u with above info. In this case, v = 0 at the top of the path.

$0-{ u }^{ 2 }=2\times 10\times 5\quad =\quad 10\quad m/s.$
Hence, the initial velocity of the ball was 10 m/s.

A body will have uniform acceleration if its

0%
speed changes at uniform rate
0%
velocity changes at uniform rate
0%
speed changes at non-uniform rate
0%
velocity remains constant

A stone falls from a balloon that is descending at a uniform rate of

$12ms^{-1}$ . The displacement of the stone from the point of release after 10sec is :-

0%
490m
0%
510m
0%
610m
0%
725m

Explanation

$S=12\times 10+\dfrac { 1 }{ 2 } \times 9.8\times { 10 }^{ 2 }\\ \ \ =120+490\\ \ \ =610\ m$

The body will speed up if ___________ .

0%
Velocity and acceleration are in same direction
0%
Velocity and acceleration are in opposite direction
0%
Velocity and acceleration are in perpendicular direction
0%
None of these

Which of the following is/are equations of uniformly accelerated motion relating the initial velocity (u), final velocity (v), time (t), acceleration (a) and displacement (S).

0%
${v^2 = u^2+2aS}$
0%
${v = u+at}$
0%
${S = ut+\dfrac{1}{2}at^2}$
0%
All the above

Explanation

The equations of motion relate to the following five quantities:
u - initial velocity
v - final velocity
a - acceleration
t - time
s - displacement
Of the above u, v, a, and s are vector quantities. As such, remember to make vectors going in one direction positive and vectors in the opposite direction negative.
Time (t) is a scalar quantity.
If any three of the five quantities are known then the other two may be calculated using the following three equations:

$v=u+at$
$s=ut+\frac { 1 }{ 2 } at^{ 2 }$
${ v }^{ 2 }{ =u }^{ 2 }+2as$

The equations of motion can only be used for an object travelling in a straight line with a constant acceleration.

When an object is accelerated:

0%
its direction must be constant
0%
its velocity is necessarily changing
0%
its velocity is constant
0%
its speed is necessarily changing

(a) Find the acceleration of the car

0%
$\, 2\, m/s^2$
0%
$4 m/s^2$
0%
$6 m/s^2$
0%
$8 m/s^2$

Explanation

Given:

Initial velocity

$u=5\ m/s$

Final velocity

$v=25\ m/s$

Time

$t=10\ s$

From the equation of motion

$v=u+at$ we get,

acceleration

$a=\dfrac{v-u}{t}=\dfrac{25-5}{10}=2\ m/s^2$

State whether given statement is True or False.
Acceleration is defined as the rate of change of velocity.

0%
True
0%
False

A particle experiences constant acceleration for

$20 s$ after starting from rest. If it travels a distance

${X}_{1}$ , in the first

$10 s$ and distance

${X}_{2}$ , in the remaining

$10 s$ , then which of the following is true ?

0%
${X}_{1} = 2{X}_{2}$
0%
${X}_{1} = {X}_{2}$
0%
${X}_{1} = 3{X}_{2}$
0%
None of these

Explanation

from eq

$s=ut+(1/2)a(t^2)$

initial velocity

$u=0 m/sec$ (particle starts from rest)

$x_1=(1/2)a(t^2)$

$x_1=(1/2)a(10^2)$

$x_1=50a$ ------------------(1)

now initial velocity for

$x_2$ is the final velocity of

$x_1$

let final velocity of

$x_1$ =initial velocity of

$x_2=v$

so,

$v=u +at$

$v=at$

$v=10a$ (since u=0 and t=10)

now

$x_2=ut+(1/2)a(t^2)$

$x_2=(10a)*10+(1/2)a(10^2)$

$x_2=100a+50a$

$x_2=150a$ -------------------------(2)

now

$\dfrac{(2)}{(1)}=\dfrac{x_1}{x_2}$

$\dfrac{x_1}{x_2}=\dfrac{50a}{150a}$

$3x_1=x_2$

Velocity-time graph of a body with uniform velocity is a straight line

0%
Parallel to x-axis
0%
Parallel to y-axis
0%
Inclined to x-axis
0%
Inclined to y-axis

Which of the following best define the acceleration of a particle:

0%
the rate of change of velocity.
0%
only experienced during a change of direction.
0%
only experienced during a change of speed.
0%
calculated by multiplying speed by velocity.
0%
always constant.

Explanation

Acceleration is defined as the rate of change of velocity.

acceleration

$a = \dfrac{change \ in \ velocity}{time \ interval}$

Hence, any change in the speed or direction would cause a change in velocity, and hence will produce acceleration.

So, the correct option is A.

Area under a speed-time graph gives :

0%
The time taken by a moving object
0%
The distance travelled by a moving object
0%
Tthe acceleration of a moving object
0%
Tthe retardation of a moving object

Identify the wrong statement about acceleration.

0%
Acceleration is rate of change of velocity per second
0%
Acceleration is rate of change of speed per second in a certain direction
0%
Acceleration is a vector quantity
0%
Unit of acceleration is $m{s}^{-1}$

Explanation

Acceleration= rate of change of velocity = change in velocity velocity / time =

$\dfrac{ms^{-1}}{s}= ms^{-2}$

Unit of acceleration is

$m{s}^{-2}$ .

Hence, option D is correct

A body executing non-uniform motion:

0%
Undergoes acceleration
0%
Never undergoes acceleration
0%
May or may not undergo acceleration
0%
Never undergoes constant acceleration

Explanation

If a body undergoes non-uniform motion, its speed changes and a body with changing speed undergoes acceleration.
This acceleration may or may not be constant.

Acceleration of the particle is given as-

$\text{acceleration} = \dfrac{ \text{ change in velocity } }{ \text{ time } }$

A body is moving vertically upwards. Its velocity changes at a constant rate from

$50 m{s}^{-1}$ to

$20 m{s}^{-1}$ in

$3 s$ . What is its acceleration?

0%
$10 m{s}^{-2}$
0%
$- 10 m{s}^{-2}$
0%
$1 m{s}^{-2}$
0%
$- 1 m{s}^{-2}$

Explanation

Initial velocity

$= 50 m{s}^{-1}$
Final velocity

$= 20 m{s}^{-1}$
Time

$= 3 s$
Acceleration =

$\dfrac{20-50}{3}$ m

${s}^{-2}$

$= - 10 m{s}^{-2}$

A scooter increases its speed from 36 km/h to 72 km/h in 10 s. What is its acceleration?

0%
${7.2 m/s^2}$
0%
${1 m /s^2}$
0%
${3.6 m/s^2}$
0%
${10 m/s^2}$

Explanation

in the given question we have
u=36 km/hr =

$36 \times \frac{5}{18}m/s=10 m/s$
v= 72 km/hr=

$72 \times \frac{5}{18}m/s=20 m/s$
t=10 s
a=?
we know the formula
v= u +at
20 = 10+ a 10
20-10 =10 a
10 = 10 a
a = 1

$m/s^{2}$

A bullet is fired from the cart vertically at the same instant cart begins to accelerate forward. Which of the following best describes the subsequent motion of the bullet?

0%
The bullet goes up and then straight back down into the cart
0%
The bullet goes up and lands in front of the cart
0%
The bullet goes up and lands behind the cart
0%
The bullet stops in the air as the cart is accelerating and "floats" until the cart stops accelerating
0%
The bullet goes up and to the right of the cart.

A car accelerates at a rate of

$5 m{s}^{-2}$ . Find the increase in its velocity in

$2 s$ .

0%
$10 m{s}^{-1}$
0%
$2.5 m{s}^{-1}$
0%
$5 m{s}^{-1}$
0%
$2 m{s}^{-1}$

Explanation

Acceleration

$= 5 m{s}^{-2}$
Time

$= 2 s$
Increase in velocity

$=$ acceleration

$\times$ time

$=(5 \times 2)m{s}^{-1}=10m{s}^{-1}$

The accelerated motion of a body changes due to change in:

0%
Speed
0%
Direction of motion.
0%
Velocity.
0%
All of the above

A car moving is with a particular velocity, which has been measured each second and the data is recorded as per the table below.
Velocity Time

$0$

$2 \ {m}/{s}$

$1 \ s$

$4 \ {m}/{s}$

$2 \ s$

$6 \ {m}/{s}$

$3 \ s$

$8 \ {m}/{s}$

$4 \ s$
Calculate the acceleration of the car ?

0%
$1\ {m}/{{s}^{2}}$
0%
$2 \ {m}/{{s}^{2}}$
0%
$4 \ {m}/{{s}^{2}}$
0%
$6 \ {m}/{{s}^{2}}$
0%
$8 \ {m}/{{s}^{2}}$

Explanation

Acceleration is defined as the ratio of change in velocity to the change in time interval.

Acceleration

$a=\dfrac{\Delta v}{\Delta t}$

The given data shows that the velocity of car changes by

$2 \ m/s$ in

$1$

$s$ each.

Thus acceleration of the car

$a = \dfrac{\Delta v}{\Delta t} = \dfrac{2}{1} =2 \ m/s^2$

In which of the following option could represent the ball increasing its speed?

A constant force acting on a body of mass of 5kg change its speed from 5

$ms^{-1}$ to 10

$ms^{-1}$ in 10 s without changing the direction of motion. The force acting on the body is

0%
1.5 N
0%
2 N
0%
2.5 N
0%
5 n

Explanation

Here, m=5 kg, u=5

$ms^{-1}$ , v=10

$ms^{-1}$ , t= 10s
Using v=u+at

$a=\dfrac{v-u}{t}=\dfrac{(10-5)ms^{-1}}{10 s}= 0.5 ms^{-2}$
As F=ma

$\therefore F=(5 kg)(0.5ms^{-2})= 2.5N$

A bird flies for 4 s with a velocity of $\left| {t - 2} \right|m/s$ in a straight
line,Where t is time in seconds.It covers a distance of

0%
2 m
0%
4 m
0%
6 m
0%
8 m

Explanation

Velocity,

$u=|t-2|$

Time,

$t=4\ s\\$

$t=0,\ u=2\ m/s\\t=2,\ u=0\ m/s\\t=4\ , u=2\ m/s$

between time

$t=0\ to\ t=2 sec$

$a=\dfrac{0-2}{2}=-1\ m/s^2$

$s_1=ut+\dfrac12 at^2\\ s_1=2\times 2+\dfrac12\times (-1)\times 4=2$

Between time

$t=2\ to\ t=4 sec$

$a=\dfrac{0-2}{2}=1\ m/s^2\\s_2=\dfrac12\times 1\times 4=2$

Thus,

Total distance;

$s_1+s_2=2+2=4\ m$

Thus, during the 4 seconds, the bird travels a distance of 4 m.

The motion of particle of mass m is given by

$y=ut+\frac{1}{2}gt^2$ . The force acting on the particle is

0%
mg
0%
$\frac{mu}{t}$
0%
2mg
0%
$\frac{2mu}{t}$

Explanation

Here

$y=ut+\dfrac{1}{2}gt^2$ .

$\therefore, v=\dfrac{dy}{dt}=u+gt$
Acceleration,

$a=\dfrac{dv}{dt}=g$
So, the net force acting on the particle is, F=ma=mg

What is the acceleration of the race car that moves at constant velocity of 300 km/hr ?

0%
73.32 $m/sec^2$ .
0%
0 $m/sec^2$
0%
63.33 $m/sec^2$ .
0%
53.33 $m/sec^2$ .

Explanation

Since the car is moving with a constant velocity, so change in the velocity of car is zero i.e.

$\Delta v = 0$
Acceleration of car

$a = \dfrac{\Delta v}{\Delta t} = 0 \ m/s^2$

Consider a train which can accelerate with an acceleration of

$20cm/s^2$ and slow down with deceleration of

$100cm/s^2$ . Find the minimum time for the train to travel between the stations 2.7km apart.

0%
90 s
0%
180 s
0%
160 s
0%
240 s

Explanation

Max. Speed , $v = u + at = 0.2 t ,$

$v = 0.2 t_1$ ------------(1)

Dec --- $v = u –at$

$0 = v – 1.0 t_2 , v = t_2$ ----(2)

${t_1} = \dfrac{{{t_2}}}{{0.2}} = 5{t_2}\,\,\,\,$

$S = ut + \dfrac{1}{2} at^2$

$S = 2700 m$
$2700 = 0.1 [ (5t_2)^2 + (t_2)^2 - \frac{1}{2} {t_2}^2]$

$r= 80 cm$

$3 { t_2}^2 = 2700$

$t_2 = 30 sec, t_1 = 150 sec$

Total time = $t_1 + t_2 = 150 + 30 = 180 sec$

The displacement of a body is given by

$s=\dfrac{1}{2}gt^{2}$ , where

$g$ is acceleration due to gravity. The velocity of the body at any time

$t$ is

0%
$\dfrac{gt^{3}}{6}$
0%
$\dfrac{gt^{2}}{6}$
0%
$gt$
0%
$\dfrac{gt}{2}$

Explanation

Given,

Displacement of body, $s=\dfrac{1}{2}a{{t}^{2}}$

Kinematic equation, $s=ut+\dfrac{1}{2}a{{t}^{2}}$

On comparing, it is clear that acceleration is equal to $a=g\,$ and initial velocity $u=0.$

Apply kinematic equation

$v=u+at$

$v=0+gt$

$v=gt$

Hence, velocity at time $t$ is $gt.$

A bullet moving with a speed of

$100ms^{-1}$ can just penetrate into two planks of equal thickness. Then the number of such planks, if speed is double will be:

0%
$6$
0%
$10$
0%
$4$
0%
$8$

Explanation

Given that,

Speed of bullet $u=100\,m/s$

Suppose that, the thickness of one plate is s

Now, from the equation of motion

${{v}^{2}}-{{u}^{2}}=-2as$

$0-{{u}^{2}}=-2as\left( s \right)$

${{u}^{2}}\propto s$

Now,

$\dfrac{v_{2}^{2}}{v_{1}^{2}}=\dfrac{{{s}_{2}}}{{{s}_{1}}}$

$\dfrac{\left( {{\left( 2\times 100 \right)}^{2}} \right)}{{{\left( 100 \right)}^{2}}}=\dfrac{{{s}_{2}}}{{{s}_{1}}}$

${{s}_{2}}=4{{s}_{1}}$

${{s}_{2}}=4\times 2s$

${{s}_{2}}=8s$

Hence, is $8$

At any instant, the velocity and acceleration of a particle moving along a straight line v and a. The speed of the particle is increasing if

0%
v > 0, a > 0
0%
v < 0, a > 0
0%
V > 0, a < 0
0%
v > 0, a = 0

The motion of a particle of mass

$m$ is describe by

$y = ut + \dfrac{1}{2} gt^2$ . Find the force acting on the particle.

0%
$F = ma$
0%
$F = mg$
0%
$F = 0$
0%
None of these

Explanation

$v=ut+\dfrac{1}{2}\times gt^2$

Differentiate y with respect to t which gives velocity as dy/dt = v

$\dfrac{dv}{dt}$ = $u + \dfrac{1}{2}g (2t)$

v=u + gt -------------------Eqn (1)

Differentiate v with respect to t which gives velocity as d2y/dt2 = dv/dt = a

$\dfrac{dv}{dt}$ = 0 + g

a = g ----------------------Eqn (2)We know that force acting on a mass m is given by F = ma

Now from Eqn (2), substitute a =g, Hence, F = mg

CBSE Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 1 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 1 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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