Question

The acceleration of a particle is increasing linearly with time t as $bt$. The particle starts from the origin with an initial velocity ${v_0}$​. The distance travelled by the particle in time t will be
A. ${v_0}t + \dfrac{1}{3}b{t^2}$
B. ${v_0}t + \dfrac{1}{3}b{t^3}$
C. ${v_0}t + \dfrac{1}{6}b{t^3}$
D. ${v_0}t + \dfrac{1}{2}b{t^2}$

Answer 1

In order to denote the position of an object we define a coordinate system and an origin as a reference point. If a body moves from one position to another position then subtracting the initial position vector from the final position vector gives us displacement. Rate of change of displacement with respect to time gives us velocity.

Formula used:
\eqalign{ & \dfrac{{dv}}{{dt}} = a \cr & \int {(a)} dt = v \cr & \int {(v)} dt = s \cr}

Complete step-by-step answer:
Rate of change of displacement with respect to time gives us the velocity and the rate of change of velocity with respect to time gives us the acceleration. Now if we are given with acceleration as a function of time and we were asked to find out the displacement then first we should get the velocity function by integrating the acceleration with in the proper limits given and then later after getting the velocity function, integrate it again with respect to time to get the displacement.
Acceleration function given is $a = bt$ and we have $\dfrac{{dv}}{{dt}} = a$
$\int {(a)} dt = v$
\eqalign{ & \Rightarrow \int {(bt)} dt = v \cr & \Rightarrow \left[ {\dfrac{{b{t^2}}}{2}} \right]_0^t = \left[ v \right]_{{v_0}}^v \cr & \Rightarrow \dfrac{{b{t^2}}}{2} = v - {v_0} \cr & \Rightarrow v = {v_0} + \dfrac{{b{t^2}}}{2} \cr}
In the above integration mechanism we integrate the function according to rules and by applying proper initial and final limits. Because it is given that the initial body is having some initial velocity ${v_0}$. So we added it to the final velocity which we get at time ‘t’
Since we had got the velocity now we integrate it to get the displacement.
$\int {(v)} dt = s$ where ‘s’ is the displacement
$\int {(v)} dt = s$
\eqalign{ & \Rightarrow \int {({v_0} + \dfrac{{b{t^2}}}{2})} dt = s \cr & \Rightarrow \left[ {{v_0}t + \dfrac{{b{t^3}}}{6}} \right]_0^t = \left[ s \right]_0^s \cr & \Rightarrow {v_0}t + \dfrac{{b{t^3}}}{6} = s - 0 \cr & \Rightarrow s = {v_0}t + \dfrac{{b{t^3}}}{6} \cr}

So, the correct answer is “Option C”.

Note: In the question it is given that initially the particle is at origin. Hence during the application of limits to find the displacement we applied the initial limit as zero. The acceleration function given here is linearly increasing with time which means velocity also will keep on increasing and there exists no time that means its direction won't change so we can consider displacement as distance.