Question

a particle is moving so that its displacement s is given as s=t^3-6t^2+3t+4 meter . its velocity at the instant when its acceleration is zero will be​

Answer 1

Answer:-

v = -6 m/s

Given :-

$S = t^3 -6t^2 +3t +4$

To find :-

It's velocity when acceleration is 0.

Solution:-

$S = t^3 -6t^2 +3t +4$

• Differentiate with respect to time.

→$\dfrac{ds}{dt}= \dfrac{d(t^3-6t^2+3t+4)}{dt}$

→$\dfrac{ds}{dt}=\dfrac{d(t^3)}{dt}-6 \dfrac{d(t^2)}{dt}+ \dfrac3{d(t)}{dt}+ \dfrac{d(4)}{dt}$

→$v = 3t^2 - 6 \times 2t + 3\times 1 + 0$

→$v = 3t^2 -12t + 3$

• Differentiate again with respect to time.

→$\dfrac{dv}{dt}= \dfrac{d(3t^2 -12t +3)}{dt}$

→$\dfrac{dv}{dt}= 3\dfrac{d(t^2)}{dt}-12\dfrac{dt}{dt}+ \dfrac{d(3)}{dt}$

→$a = 6t -12$

• at a = 0 m/s²

→$0 = 6t -12$

→$12 = 6t$

→$t = \dfrac{12}{6}$

→$t = 2s$

• at t = 2s

The velocity will be :-

→$v = 3t^2 -12 t + 3$

→$v = 3 (2)^2 -12 \times 2 +3$

→$v = 12 -24 +3$

→$v = 15 -24$

→$v = -9 m/s$

hence,

The velocity will be -9 m/s.