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. ${V_0}t + \dfrac{1}{2}b{t^2}$
C. ${V_0}t + \dfrac{1}{6}b{t^3}$
D. ${V_0}t + \dfrac{1}{2}b{t^3}$

Answer 1

In order to solve this problem we need to use the formula of acceleration and then integrate this we get the value then after we use the formula of velocity then we integrate. So that we can calculate the distance travelled by the particle.

Complete step by step answer:
From the given data:
$a = bt$
WKT, a=change in velocity/change in time
$a = \dfrac{{dv}}{{dt}} \\\ \dfrac{{dv}}{{dt}} = bt \\\$
Now we will integrate the above we get V
$v = \dfrac{{b{t^2}}}{2} + c$
At t=0 sec, $v = {v_0}$
Therefore
${v_0} = c \\\ v = \dfrac{{b{t^2}}}{2} + {v_0} \\\$
From the velocity formula,
V=Change in position/change in time
$v = \dfrac{{ds}}{{dt}} \\\ v = \dfrac{{b{t^2}}}{2} + {v_0} \\\ \\\$
Equate the above equation,
$ds = \left( {\dfrac{{b{t^2}}}{2} + {v_0}} \right)dt$
Now we will integrate the above we get S,
$s = \dfrac{{b{t^3}}}{6} + {v_0}t + {c} \\\ t = 0,s = 0,{c’} = 0 \\\ s = \dfrac{{b{t^3}}}{6} + {v_0}t \\\$

Hence the correct option is C.

Note: In this type of problem, integration with limit is used to avoid the integration constant occurrence. It consists of upper and lower limits for the velocity of a particle V.