E) A particle moves in a straight line such that itsacceleration a var

1.	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» 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.

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»

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.

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