Question

A body moves with initial velocity $10m{s^{ - 1}}$. If it covers a distance of $20m$ in $2s$,then acceleration of the body is
(A) Zero
(B) $10m{s^{ - 2}}$
(C) $5m{s^{ - 2}}$
(D) $2m{s^{ - 2}}$

Answer 1

Hint: We are going to use the equations of one-dimensional motion to solve this problem. One-dimensional motion describes objects moving in a straight line. An example is a car moving down a road.
Velocity is the rate of change of displacement and acceleration can be defined as the rate of change of velocity. Since it has both direction and magnitude it is a vector quantity.

Complete step by step solution:
We know that,
${\text{s = ut + }}\dfrac{1}{2}{\text{a}}{{\text{t}}^2}$
Where s is the distance
U is the initial velocity
A is the acceleration.
T is the time taken.
We are given that,
$s = 20m$ $u = 10m{s^{ - 1}}$ $t = 2s$
Substituting these values in the equation gives
$ \Rightarrow 20 = 10 \times 2 + \dfrac{1}{2} \times a \times {2^2}$
$ \Rightarrow 20 = 20 + 2a$
$ \Rightarrow a = 0$

Hence the correct answer is Option A.

Additional Information:
Uniform acceleration: It can be defined as if the velocity of an object is increasing at constant rate then the object is said to have uniform acceleration.
Average acceleration: It can be defined as the total change in velocity interval to the total time taken in a particular time interval.
Instantaneous acceleration: It can be defined as the ratio to change in velocity during a given time interval such that the time interval goes to zero.

Note:
Angular acceleration: It is defined as the rate of change of angular velocity It is also known as rotational acceleration. If there is an increase in angular velocity clockwise then angular acceleration points at a direction away from the observer. If there is decrease in angular velocity clockwise then angular acceleration points at a direction towards the observer.