Question

The position of a particle is r⃗ (t)=(3.0t2iˆ+5.0jˆ−6.0tkˆ)m. (a) Determine its velocity and acceleration as functions of time. (b) What are its velocity and acceleration at time t = 0?

Answer 1

Explanation:

It is given that,

The position of a particle is given by :

$r(t)=(3t^2i+5j-6tk)\ m$

(a) Velocity of a particle is given by :

$v=\dfrac{dr(t)}{dt}$

Putting values,

$v=\dfrac{d}{dt}(3t^2+5-6t)\\\\v=(6ti-6k)\ m/s$

The acceleration of the particle is given by :

$a=\dfrac{dv}{dt}\\\\a=\dfrac{d}{dt}(6t-6)\\\\a=6i\ m/s^2$

(b) At t = 0,

Velocity, v = 6k m/s

Acceleration, a = 6i m/s²