Question

A particle starts moving and its displacement after $t$ seconds is given in meter by the relation $x=5+4t+3t^2$. Calculate the magnitude of its
a. Initial velocity
b. Velocity at $t=3s$
c. Acceleration.

Answer 1

Velocity $=$ derivative of displacement
Acceleration $=$ derivative of velocity
a. Initial velocity means velocity at $t=0$. Now, $v=dx/dt$,
i.e., $v=4+6t$. So initial velocity $=4+6\times 0=4m{s}^{-1}$.
b. Velocity at $t=3s:v=4+6\times 3=22m{s}^{-1}$.
c. Acceleration, at any time $t:a=\frac{dv}{dt}=6m{s}^{-2}$. This is a constant acceleration.