Question

e) A particle moves in a straight line such that itsacceleration a varies with speed vasa =-2(v )^1/2 m/s^2, where v is measured in m/s. Ifinitial velocity is 4 m/s, then time taken by theparticle to come to stop is​

Answer 1

Answer:

2 seconds

Explanation:

given,

$a = 2{v}^{ \frac{1}{2} }$

$u = 4 \: \frac{m}{ {s}^{2} }$

as we know

$a = \frac{dv}{dt}$

$\frac{dv}{dt} = 2 {v}^{ \frac{1}{2} } \\ \\ \frac{dv}{ {v}^{ \frac{1}{2} } } = 2dt$

now integrating both the sides with limit 4 to 0 for velocity and 0 to t for time.

we get

$- 2 {v}^{ \frac{1}{2} } = 2t$

now put the limits in the result obtained

we get

$( 2) \times 2 = 2 \times t \\ 4 = 2t \\ \frac{ 4}{2} = t \\ or \\ t = 2$

hence the particle will stop in 2 seconds.