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Equation of motion - class-XI

Description: equation of motion
Number of Questions: 19
Created by:
Tags: force and motion motion motion in a straight line equations of motion motion in one dimension accelerated motion physics mechanics
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Derivation of second equation of motion is:

  1. $d\theta= w d2t$

  2. $d\theta= w dt$

  3. $d\theta= w d3t$

  4. $d\theta= w dt^{2}$


Correct Option: B
Explanation:

$S= ut+\frac{1}{2} at^2$
$\theta = w _0t +\frac{1}{2}\alpha t^2$
$w= \frac{d\theta}{dt}$
$d\theta= w dt$

The distance $x$ covered in time $t$ by a body having initial velocity ${ v } _{ 0 }$ and having constant acceleration $a$ is given by $x={ v } _{ 0 }t+1/2a{ t }^{ 2 }$. This result follows from :

  1. newton's Ist law

  2. newton's IInd law

  3. newton's IIIrd law

  4. none of the above


Correct Option: D
Explanation:

The given equation is the fundamental equation of kinematics. So it can not follow from any Newton's law of motion.
Ans:(D)

A ladder of length 10 m and mass 20 kg (and with uniform mass distribution) leans against a slippery vertical wall. The ladder makes an angle of $30^{\circ}$ with respect to the vertical. Friction between the ladder and the ground prevents it from sliding downwards. What is the magnitude of the force exerted on the ladder by the wall?
$[Take \sqrt { 3 } =1.732;g=10{ m/s }^{ 2 }]$

  1. $0 N$

  2. $0.58 N$

  3. $58 N$

  4. $5.8 N$


Correct Option: C

For a body moving with an initial velocity $u$ and uniform acceleration $a$. Find the displacement of the body in time t.

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

  2. $s=u+at$

  3. $s=ut-\dfrac{1}{2}at^2$

  4. $s=ut+at^2$


Correct Option: A
Explanation:

From the second equation of motion, if u = initial velocity , a= acceleration and t= time then

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

How is the distance related with time for the motion under uniform acceleration such as the motion of a freely falling body starting from rest?

  1. $S \propto t^2$

  2. $S \propto t$

  3. $S \propto \dfrac{1}{t^2}$

  4. $S \propto \dfrac{1}{t}$


Correct Option: A
Explanation:

For a free falling body, initial velocity, $u = 0 m/s $ and acceleration due to gravity, $a = g$.
Putting the values of $u$ and $a$ in the equation of motion, $S = ut+\dfrac{1}{2}at^2 $
We get, $S = \dfrac{1}{2}gt^2$
Therefore, $S \propto t^2$

The correct equation of motion is :

  1. $v=u+aS$

  2. $v=ut+a$

  3. $S= ut +\frac{1}{2} at$

  4. $v = u + at$


Correct Option: D
Explanation:

Three equations of motion are as follows:

$v=u+at$

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

$v^2 = u^2+2aS$

In the equation of motion, $S = ut + 1/2 at^2$, S stands for

  1. displacement in t seconds

  2. maximum height reached

  3. displacement in the $t^{th}$ second

  4. none of these


Correct Option: A
Explanation:

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

where 
S= Distance traveled in the t seconds
u= Initial velocity of the motion 
a= acceleration or retardation of the motion 

Gradient of line of displacement-time graph gives the :

  1. acceleration

  2. velocity

  3. displacement

  4. none of these


Correct Option: B
Explanation:
Slope of line of displacement time graph$=\cfrac { ds }{ dt } $
The rate of change of position i.e., rate of change of displacement with time is called velocity

Gradient of line of velocity-time graph gives :

  1. distance

  2. time

  3. velocity

  4. acceleration


Correct Option: D
Explanation:
Gradient of line is basically a slope of line
Slope$=\cfrac { \triangle y }{ \triangle x } $
Here, Slope $=\cfrac { d\upsilon  }{ dt } $
As we all know acceleration is the rate of change of velocity with time
$\therefore $ acceleration $=\cfrac { d\upsilon  }{ dt } $

Two trains A and B each of length 400m Are moving on two parallel tracks The same direction (while A is ahead of B) With same speed 72 km / h.The driver of B decides to overtake And accelerate by $1m/{ s }^{ 2 }$. If after 50 seconds B Just brushes past A, Calculate the original distance between A and B


  1. $850m$

  2. $1000m$

  3. $1250m$

  4. $2250m$


Correct Option: A

At an airport, a bored child starts to walk backwards on a moving platform. The child accelerates relative to the platform with $a =  - 0.5m/{s^2}$ relative to the platform. The platform moves with a constant speed $v =  + 1.0m/s$ relative to the stationary floor. In $4.0$ seconds, how much will the child have been displaced relative to the floor?

  1. $8m$

  2. $0m$

  3. $3m$

  4. $4m$


Correct Option: B
Explanation:
initial vel = vel of platform $=1\ m/s$
acceleration wrt floor $=$ acceleration of him wrt platform +acc. of platform
$=\dfrac {-1}{2}m/s^2+0=\dfrac {-1}{2}m/s^2$
So, for finding displacement of him in us,
Applying second equation of motion
$x=ut +\dfrac {at^2}{2}$
$(1)(4)^2-\left (\dfrac {1}{4}\right) \dfrac {(4)^2}{2}$
$x=4-\dfrac {(4)^2}{4}=0\ m$

An open lift is moving upwards with velocity $10m/s$. It has an upward acceleration of $2m/{s^2}$. A ball is projected upwards with upwards with velocity $20m/s$ relative to the ground. Find the time when the ball again meets the lift.

  1. $\frac{5}{3}s$

  2. $\frac{4}{3}s$

  3. $\frac{3}{4}s$

  4. $\frac{1}{4}s$


Correct Option: A
Explanation:
Let the time of Height of the ball be $t$. Then. with respect to the lift.
$V _{B/L}= V _{B}-V _{L}= 20-10= 10\ m/s$
$a _{B/L}=a _{B}-a _{L}=(-10)-(2)=-12\ m/s^{2}$
To find $t$, applying second $eq^{n}$ to motion 
$h =u (t)+ at^{2}$
$\Rightarrow o= (10)(t)+ \dfrac{(-12)t^{2}}{2}$
$\Rightarrow bt^{2}=10 t$
$\Rightarrow t = 5/3^{3}$
Option $-A$ is correct. 

An elevator car whose floor-to-ceiling distance is equal to 2.7m starts ascending with a constant acceleration $1.2m/{s^2};2.0s$ after the start a bolt begins falling from the ceiling of the car. Find the displacement covered by the bolt during the free fall in the reference frame fixed to the elevator shaft.

  1. 0.7 m

  2. 1.7 m

  3. 2.7 m

  4. 3.7 m


Correct Option: A
Explanation:
Let the total time of free fall be $t$
Then at $t=2s$
$4 _{B}=2. \dfrac{1}{m}/s$          $4 _{C}=2.\dfrac{1}{m}/s$
$a _{C}=1.2\ m/s^{2}$              $a _{B}=-10\ m/s^{2}$
Wrt car.
distance travelled by bolt $= 27\ m$
Then 
$-2.7\ m=(4 _{B/C})t+a _{B/C}\left(\dfrac{t^{2}}{2}\right)$
$\Rightarrow -2.7= (2.4-2.4)t (\dfrac{-10-12}{2}) t^{2}$
$\Rightarrow 2.7= \dfrac{11.2}{2}t^{2}$
$\Rightarrow t^{2}=\left(\dfrac{2.7}{11.2}\times 2\right)^{-11}=0.482\ s^{2}$
Note: $4 _{B}=4 _{C}$ till $t=2.3$ before the both spots to fall. 
$4 _{B}=4 _{C}=a _{C}t=(1.2)(2)=2.4 m/s$
let the displacement of the bolt be $h$ Then, 
$h=4 _{B}(t)+4 _{\dfrac{B}{2}}t^{2}$
$h=(2.4)\sqrt{0.482}\dfrac{-(10)}{2}(0.482)$
$h=-0.743$
$h=\approx -0.7\ m$
$|h|\approx 0.7\ m$
Option $A$ is correct

A boy sitting on the top most berth in the compartment of a train which is just going to stop on the railway station, drops an apple aiming at the open hand of his brother situated vertically below his hands at a distance of about 2 m. The apple will fall

  1. In the hand of his brother

  2. Slightly away from the hand of his brother in the direction of the motion of the train.

  3. Slightly away from the hands of his brother in the direction opposite to the direction of the motion of the train.

  4. None of these


Correct Option: B
Explanation:

As both the boy and his brother are moving in same frame,and train is going to be slow down when the train will de-accelerate it will impart an inertia force in opposite direction but at the same time the boy will drop an apple from 2m height . when the train will exert inertia force on boy due to action reaction boy will also expert an force in forward direction due to which he will not able to put the apple in the hand of his brother and it will fall in forward direction.

The distance covered by a body moving alone-X-axis with initial velocity 'u' and uniform acceleration 'a' is given by $\displaystyle x=ut+\frac{1}{2}at^{2}$. This result is a consequence of

  1. Newton's Ist law

  2. Newton's IInd law

  3. Newton IIIrd law

  4. None of the above


Correct Option: D
Explanation:

There is no relation between NLM and given equation

A cart begins from rest at the top of a long incline and rolls with a constant acceleration of $2m/s^{2}$. How far has the cart moved along the incline after rolling for $3$ seconds?

  1. $3$ meters

  2. $6$ meters

  3. $9$ meters

  4. $18$ meters


Correct Option: C
Explanation:

Initial velocity of cart      $u =0  m/s$

Acceleration of the cart       $a = 2$ $m/s^2$
Distance covered by cart in 3 seconds       $S = ut + \dfrac{1}{2}at^2$                  where  $t =3$ seconds
$\therefore$    $S = 0 + \dfrac{1}{2 } \times 2 \times 3^2  = 9$  meters

A car was travelling at a speed of 10 m/s. After the brakes were applied, it start to decelerate at a rate of 2 $m/s^2$ and finally came to rest. Find the distance traveled by the car before coming to rest.

  1. 10 m

  2. 15 m

  3. 25 m

  4. 35 m


Correct Option: C
Explanation:
Initial speed of the car   $u = 10 \ m/s$
Final speed of the car   $v =  0 \ m/s$
Acceleration (or  retardation) of the car   $a = -2 \ m/s$
Using   $v^2 - u^2 = 2aS$
$\therefore$   $0 - 10^2 = 2(-2)S$
$\implies \ S = \dfrac{100}{2\times 2}  = 25 \ m$

The initial velocity of a particle (at $t = 0$) is $u$ and the acceleration by $f$ of particle at time $t$ is given $f=at$. Where $a$ is a constant which of the following relation for velocity $v$ of particle after time $t$ is turn 

  1. $v\, =\, u\, +\, at^2$

  2. $v\, =\, u\, +\, \displaystyle \frac{at^2}{2}$

  3. $v\, =\, u\, +\, at$

  4. $v\, =\, u$


Correct Option: C
Explanation:

We know that,

$\dfrac{dv}{dt} = a$
Hence,
$dv = a dt$
Integrating this and applying the proper limits.
$v  - u = at$
$\implies v=u+at$
Hence, option C is correct.

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