Question

A cat running with constant acceleration in one direction is at rest initially and has velocity v(t)=1 m/s, 2 seconds later.
At t=4 sec, how far away from the starting point will the cat be?
A. 0 m
B. 1 m
C. 2 m
D. 4 m
E. 8 m

Answer 1

In order to get relation between velocity and acceleration and also between time and distance, we have well derived equations for kinematics which could be used to calculate different parameters related to kinematics.

Formula used:
$v=u+at, \ s = ut+\dfrac12 gt^2$

Complete answer:
Here we are given that the body was initially at rest. Thus initial velocity (u) will be zero. We are supposed to find the distance traveled in 2 seconds. But for that, we need acceleration of the body. Thus to obtain acceleration, we have to use:
$v = u+at$
Given at t = 2 sec, v = 1m/s.
Thus $1 = 0 + a(2)$
$\implies a = \dfrac12 m/s^2$
Now, in order to find the distance travelled between this time interval of 4 sec, we have to apply:
$s = ut+\dfrac12 at^2$
$s = 0\times 4+\dfrac12 \left(\dfrac12\right)4^2 = 4m$
Thus, option D. is correct.

Additional information:
In the statement of this question, we are given that the acceleration is constant for the cat. But in case the acceleration is variable, we gave to proceed with the differential equation of motion, i.e. $v = \dfrac{dx}{dt} \ and \ a = \dfrac{dv}{dt}$ and integrate the expression under the given limits. Note that this method will be valid irrespective of acceleration to be constant or variable.

Note:
One should note here that the reason of using the equation$v=u+at$, instead of $s=ut+\dfrac 12 at^2$ or $v^2-u^2=2as$ is that there is nowhere in the question we are dealing with distances or displacement. One can use them to relate velocity and displacement, when time is not given ($v^2-u^2=2as$) and when we are asked to find displacement in given time and final velocity is not given ($s=ut+\dfrac 12 at^2$).
Chance of mistake is that students should note that these equations are valid only for constant acceleration.