Question

The velocity time relation of a body starting from rest is given by $v = k^{2} t ^{\dfrac{3}{2}}$, where $k = \sqrt{2} \left(m^{\dfrac{1}{2}} s^{\dfrac{-5}{2}} \right)$. What is the distance traversed in $4$ seconds?
a) $2.00 m$
b) $2.56 m$
c) $25.6 m$
d) $256 m$

Answer 1

Hint: We have given the velocity – time relation, first after putting the value of k; we find the value of velocity. Then, we write velocity in terms of differential form. Velocity is the rate of change of distance. After integration, we get the distance in terms of time. Then, we will put the value of time and get distance at required time.

Complete step-by-step solution:
Given: the velocity – time relation is -
$v = k^{2} t ^{\dfrac{3}{2}}$
where, $k = \sqrt{2} \left(m^{\dfrac{1}{2}} s^{\dfrac{-5}{2}} \right)$
Here, m is the meter.
s is the second.
Now put the value of k in the velocity – time relation.
$v = \left( \sqrt{2} \right)^{2} t ^{\dfrac{3}{2}}$
It gives,
$v = 2 t ^{\dfrac{3}{2}}$
As we know,
$v = \dfrac{ds}{dt}$
Now, evaluate the value of velocity.
 $\dfrac{ds}{dt} = 2 t ^{\dfrac{3}{2}}$
Now, separate the variables and integrate the time from $0$ to $4$ to find the value of distance.
$ \int {ds}=\int^{4}_{0} 2 t ^{\dfrac{3}{2}} dt$
After integration, we get,
$s = 2 \left|\dfrac{t^{\dfrac{5}{2}}}{\dfrac{5}{2}} \right|^{4}_{0}$
Now put the limits and then
$s = \dfrac{4}{5} \times 4^{\dfrac{5}{2}}$
We get the value of distance,
$s = 25.6 m$
The distance traversed in $4$ seconds is $25.6 m$.
Option (c) is correct.

Note: Velocity-Time graph is significant while learning about the movement of any object, the graph represents the object's acceleration. The curve of the velocity-time graph gives acceleration as a function of time. As the acceleration is continuously rising with time, the number or magnitude of the slope will also continuously rise with time.