Question

e) A particle moves in a straight line such that its
acceleration a varies with speed vas
a =-2(v )^1/2 m/s^2, where v is measured in m/s. If
initial velocity is 4 m/s, then time taken by the
particle to come to stop is​

Answer 1

Given:

Particle moves in a straight line such that it's acceleration "a" varies with velocity as

$a = - 2 \sqrt{v}$

Initial velocity is 4 m/s

To find:

Time taken by particle to stop.

Calculation:

Acceleration is defined as the instantaneous rate of change of velocity with respect to time. So , any acceleration vs velocity function can be integrated to get velocity in terms of time.

$\therefore \: \: a = - 2 \sqrt{v}$

$= > \dfrac{dv}{dt} = - 2 \sqrt{v}$

$= > \dfrac{dv}{ \sqrt{v} } = - 2 \: dt$

Integrating on both sides :

$\displaystyle = > \int \dfrac{dv}{ \sqrt{v} } = - 2 \: \int dt$

Putting the limits:

$\displaystyle = > \int_{4}^{0} \dfrac{dv}{ \sqrt{v} } = - 2 \: \int_{0}^{t} dt$

$= > \bigg \{2 {v}^{ \frac{1}{2} } \bigg \}_{4}^{0} = - 2 \bigg \{t \bigg \}_{0}^{t}$

Reversing the limits on left side :

$= > \bigg \{2 {v}^{ \frac{1}{2} } \bigg \}_{0}^{4} = 2 \bigg \{t \bigg \}_{0}^{t}$

$= > 2 \times ( {4}^{ \frac{1}{2} } - 0)= 2 t$

$= > 2t = 2 \times 2$

$= > t = 2 \: sec$

So , final answer is :

Car will stop after 2 seconds.