CBSE Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 9 - MCQExams.com

Apply equation of kinematic

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

$$ {{t}^{2}}+\dfrac{2ut}{a}-\dfrac{2s}{a}=0 $$

$$ {{t}^{2}}+pt-ks=0 $$

Where, $$p=\dfrac{2u}{a}\,\,and\,\,k=\dfrac{2}{a}$$

$$t=\dfrac{-p\pm \sqrt{{{p}^{2}}+ks}}{2}$$

Hence, time is directly proportional to root of displacement.

Let assume initial velocity is $$u$$ and constant acceleration $$a$$. using    $$s$$=$$ut$$ + $$(0.5)at^2$$, 

for first $$5$$ second

$$10$$ = $$5u$$ + $$(0.5)a(25)$$

for next 3 seconds, $$t = 8$$ and $$s = 20$$

$$20 = 8u + (0.5)a(64)$$

Solving,

$$a = {\dfrac{1}{3}}$$ and $$u = {\dfrac{7}{6}}$$

Now for next $$2$$ seconds, $$t = 10$$ and $$S'$$:

$$S' = (10){\dfrac{7}{6} +(0.5){\dfrac{1}{3}}(100)}$$

$$S'$$ = $${\dfrac{170}{6}}$$

Now distance travelled in $$2$$ sec = $${\dfrac{170}{6}} - {20} = {\dfrac{50}{6}}$$ meter

Given that,

Initial velocity $$u=20\,m/s$$

Final velocity $$v=0$$

Now, from equation of motion

  $$ v=u-gt $$

 $$ 0=20-10t $$

 $$ {{t}_{1}}=\dfrac{20}{10} $$

 $$ {{t}_{1}}=2\,s $$

Total time

 $$ t=2{{t}_{1}} $$

 $$ t=2\times 2 $$

 $$ t=4\,s $$

Hence, the time is $$4\ s$$

Given,

Initial velocity $$=u$$

After $$2\ cm\,$$penetration, velocity $$v=\dfrac{u}{2}$$

Apply equation of kinematic

 $$ {{v}^{2}}-{{u}^{2}}=2as $$

 $$ {{\left( \dfrac{u}{2} \right)}^{2}}-{{u}^{2}}=2a\times 2 $$

 $$ a=\dfrac{-3{{u}^{2}}}{16} $$

When, Final velocity $${{v}_{1}}=\dfrac{u}{4}$$

Apply kinematic equation of motion

 $$ v_{1}^{2}-{{u}^{2}}=2a{{s}_{1}} $$

 $$ {{\left( \dfrac{u}{4} \right)}^{2}}-{{u}^{2}}=2\left( \dfrac{-3{{u}^{2}}}{16} \right){{s}_{1}} $$

 $$ {{s}_{1}}=2.5\ cm $$

Hence, for final velocity to be one fourth of initial velocity, body penetrate $$0.5\ cm$$ more in wood.

Given,

Horizontal velocity, $${{v}_{x}}=20\,m{{s}^{-1}}$$

Vertical initial velocity, $${{u}_{y}}=25\,m{{s}^{-1}}$$

Acceleration of gravity, $${{a}_{y}}=-10\,m{{s}^{-2}}$$

Apply equation of kinematic

$$ {{v}_{y}}={{u}_{y}}+{{a}_{y}}t $$

$$ {{v}_{y}}=25-10\times 2=5\,m{{s}^{-1}} $$

Hence, velocity $$20\,m{{s}^{-1}}$$ horizontal and $$5\,m{{s}^{-1}}$$ upward

Given that,

Mass $$m=2\,kg$$

Initial velocity $$u=0\,m/s$$

Final velocity $$v=20\,m/s$$

Time $$t=4\,s$$

Now, the acceleration is

  $$ v=u+at $$

 $$ a=\dfrac{v-u}{t} $$

 $$ a=\dfrac{20}{4} $$

 $$ a=5\,m/{{s}^{2}} $$

Now, the force is

  $$ F=ma $$

 $$ F=2\times 5 $$

 $$ F=10\,N $$

Now, the displacement is

  $$ s=ut+\dfrac{1}{2}a{{t}^{2}} $$

 $$ s=0+\dfrac{1}{2}\times 5\times 16 $$

 $$ s=40\,m $$

Now, the work done is

  $$ W=F\centerdot s $$

 $$ W=10\times 40 $$

 $$ W=400\,J $$

Now, the power is

  $$ P=\dfrac{W}{t} $$

 $$ P=\dfrac{400}{2} $$

 $$ P=200\,watt $$

Hence, the power is $$200\ watt$$ 

Given that,

Acceleration $$a=2\,m/{{s}^{2}}$$

Retardation $$a=-3\,m/{{s}^{2}}$$

Time $$t=10\,s$$

Now, from equation of motion

The maximum velocity attained be v at t  

  $$ v=u+at $$

 $$ v=0+2\times t $$

 $$ v=2t\,m/s.....(I) $$

Now, again it reaches velocity 0 after 10 s with retardation

  $$ v=u+at $$

 $$ 0=v-3\left( 10-t \right) $$

Now from equation (I)

  $$ 2t-3\left( 10-t \right)=0 $$

 $$ 2t-30+3t=0 $$

 $$ 5t=30 $$

 $$ t=6\,s $$

Now, put the value of t in equation (I)

  $$ v=2t $$

 $$ v=2\times 6 $$

 $$ v=12\,m/s $$

Hence, the maximum speed is $$12\ m/s$$

Initial velocity $$u=20\,m/s$$

Final velocity $$v=0\,m/s$$

Acceleration $$a=-10\,m/{{s}^{2}}$$

Now, the time by ball reach its maximum height

From equation of motion

  $$ v=u+at $$

 $$ 0=20-10t $$

 $$ t=2\,s $$

So, the time taken by the same ball to return to the hands of the juggler is  

  $$ =\dfrac{2u}{g} $$

 $$ =\dfrac{2\times 20}{10} $$

 $$ =4\,s $$

So, he is throwing the balls after 1 sec each,

Let at some instant he throws ball number 4.

 Now, Before 1 s of throwing it, he throws ball 3.

So, the height of ball 3 is

  $$ {{h}_{3}}=20\times 1-\frac{1}{2}\times 10\times {{\left( 1 \right)}^{2}} $$

 $$ {{h}_{3}}=15\,m $$

Before 2s, he throws ball 2, so the height of ball 2 is

  $$ {{h}_{2}}=20\times 2-\frac{1}{2}\times 10\times 4 $$

 $$ {{h}_{2}}=20\,m $$

Before 3s, he throws ball 1, so the height of ball 1 is

  $$ {{h}_{1}}=20\times 3-\frac{1}{2}\times 10\times 3\times 3 $$

 $$ {{h}_{3}}=15\,m $$

Hence, the height of balls in air from the ground will be $$15\ m$$, $$20\ m$$, $$15\ m$$ and $$0\ m$$

Hint: Apply kinematical equations.

Correct Option: Option A

Explanation for correct answer:

Step 1: Find the length of train in terms of velocity and acceleration.

• The displacement of the train is given by:

Given that,

Acceleration, $$a=\frac{dv}{dt}=4-2t$$

At t = 0 s , $$a=4m/{{s}^{2}}$$

At t = 1 s, $$a=2\,m/{{s}^{2}}$$

At t = 2 s, $$a=0$$

At t = 3 s, $$a=4-6=-2\,m/{{s}^{2}}$$

Hence, the acceleration of the body is initially positive but after some time it becomes negative.

Given that,

Height $$h=5\,m$$

$$g=10\,m/{{s}^{2}}$$

Now, for first drop

  $$ {{s}_{1}}=ut+\dfrac{1}{2}g{{t}^{2}} $$

 $$ 5=0+5{{t}^{2}} $$

 $$ t=1\,\sec  $$

It means that the third drop leaves after one second of the first drop. Or, each drop leaves after every $$0.5\ sec$$

Now, distance covered by the second drop in $$0.5\ s$$

  $$ {{s}_{2}}=ut+\dfrac{1}{2}g{{t}^{2}} $$

 $$ s=0+\dfrac{1}{2}\times 10\times 0.5\times 0.5 $$

 $$ s=1.25\,m $$

 Therefore, the distance of the second drop above the ground

  $$ s={{s}_{1}}-{{s}_{2}} $$

 $$ s=5-1.25 $$

 $$ s=3.75\,m $$

Hence, this is the required solution 

Given that,

$$ a=b{{t}^{n}} $$

$$ \dfrac{dv}{dt}=b{{t}^{n}} $$

$$ dv=b{{t}^{n}}dt $$

$$ v=b\dfrac{{{t}^{n+1}}}{n+1}........(1) $$

$$ \dfrac{dr}{dt}=b\dfrac{{{t}^{n+1}}}{n+1} $$

$$ dr=\dfrac{b}{n+1}\int{{{t}^{n+1}}dt} $$

$$ r=\dfrac{b}{n+1}\dfrac{{{t}^{n+2}}}{n+2} $$

$$ r=\dfrac{Vt}{n+2}\,\,(from\,1) $$

$$ at\,\,t=\,1\,s $$

$$ V=(n+2)r $$

The formula for time of flight is

$$T=\dfrac{2u\sin \theta }{g}$$

It is given that time is 2 seconds. So,

$$ 2=\dfrac{2u\sin \theta }{g} $$

$$ g=u\sin \theta  $$

Squaring both sides

$${{g}^{2}}={{u}^{2}}{{\sin }^{2}}\theta ........(1)$$

Height travelled is

$$ H=\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g} $$

$$ H=\dfrac{{{g}^{2}}}{2g}\,\,\,\,(from\,\,1) $$

$$ H=\dfrac{g}{2}=5\,m $$

Given,

Mass of ship, $$M=3\times {{10}^{7}}\,kg$$

Force, $$F=5\times {{10}^{4}}\,N$$

Distance, $$s=3\,m$$

Acceleration, $$a=\dfrac{F}{M}=\dfrac{5\times {{10}^{4}}}{3\times {{10}^{7}}}=\dfrac{1}{600\,}\,m{{s}^{-2}}$$

Apply kinematic equation of motion

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

$$ v=\sqrt{2as}=\sqrt{2\times \dfrac{1}{600}\times 3}=0.1\,m{{s}^{-1}} $$

Hence, velocity of ship $$0.1\,m{{s}^{-1}}$$

Given that,

Distance $$s=45\,m$$

Initial velocity $$u=0$$

We know that,

By using equation of motion

  $$ {{v}^{2}}={{u}^{2}}+2as $$

 $$ {{v}^{2}}=0+2\times 9.8\times 45 $$

 $$ {{v}^{2}}=882\,m/s $$

 $$ v=29.7\,m/s $$

Hence, the minimum speed is $$29.7\ m/s$$

It is given that, the body reaches the maximum height in 6s. Let u be the initial velocity. At maximum height the velocity is zero. Using the equation of motion as

$$ v=u+at $$

$$ 0=u-(10)(6) $$

$$ u=60\,m/s $$

Distance travelled in n^th second is given by:

$${{S}_{n}}=u+\left( \dfrac{a}{2} \right)\left( 2n-1 \right)$$

Displacement in 1^st second is $${{S}_{1}}=60-\left( \dfrac{10}{2} \right)(2-1)=55\,m$$

Displacement in 7^th second is $${{S}_{2}}=60-\left( \dfrac{10}{2} \right)(14-1)=-5\,m$$

So, $$\dfrac{{{S}_{1}}}{{{S}_{2}}}=\dfrac{55}{5}=\dfrac{11}{1}$$

The ratio is $$11:1$$.