Question

The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity Vo. The distance travelled by the particle in time t will be ​

Answer 1

We are given the acceleration and the initial velocity.

Acceleration, $a=bt$

Initial velocity, $u=v_0$

We know the second kinematic equation,

$s=ut+\dfrac {1}{2}at^2$

By this, we get that, the displacement of the particle in a time t will be,

$s=v_0t+\dfrac {1}{2}(bt)t^2\\\\\\s=v_0t+\dfrac {1}{2}bt^3$

But this answer is absolutely wrong! Because kinematic equations are applicable only in the case of constant acceleration!

In the question, since the acceleration is increasing linearly with time, it is no more a constant. So we can't apply second kinematic equation directly. In such cases integration is better.

We have, acceleration,

$a=bt$

But we know that it is the first derivative of the velocity, i.e.,

$\dfrac {dv}{dt}=bt$

Then,

$\displaystyle dv=bt\ dt\\\\\\v=\int bt\ dt\\\\\\v=\dfrac {1}{2}bt^2+c$

Here, $c=v_0.$ Then,

$v=v_0+\dfrac {1}{2}bt^2$

But, we also know that the velocity is the first derivative of the displacement, i.e.,

$\dfrac {ds}{dt}=v_0+\dfrac {1}{2}bt^2\\\\\\ds=\left (v_0+\dfrac {1}{2}bt^2\right)\ dt\\\\\\s=\int\left (v_0+\dfrac {1}{2}bt^2\right)\ dt\\\\\\s=\int v_0\ dt+\dfrac {b}{2}\int t^2\ dt\\\\\\s=v_0t+\dfrac {1}{2}b\left (\dfrac {t^3}{3}\right)\\\\\\\large\boxed {s=v_0t+\dfrac {1}{6}bt^3}$

This is the actual displacement.