Question

A particle moves according to the law s =t³ - 6t² + 9t + 15, find the velocity when t = 0.

Answer 1

On differentiating, we get dx/dt=3t^2-12t+9
V=3t^2-12t+9
At t=0, v=3(0)-12(0)+9
=9m/s

Answer 2

Answer:

Velocity = 9

Step-by-step explanation:

As per the given question,

Movement of particle with respect to time is given:

$S=t^{3}-6t^{2}+9t+15$

For finding the velocity, v

We need to differentiate the given equation with respect to time, t

Therefore,

$S=t^{3}-6t^{2}+9t+15$

$\frac{ds}{dt} =3t^{2}-12t+9$

$Velocity =3t^{2}-12t+9$

Now on putting the value of t = 0, we get

$Velocity =3(0)^{2}-12(0)+9$

∴ Velocity = 9

Hence,  the required velocity = 9.