Question

A particle starts moving and its displace-ment 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+6xx0=4ms^-1`.
b. Velocity at `t=3s` : `v=4+6xx3=22ms^-1`.
c. Acceleration, at any time `t: a=(dv)/(dt)=6ms^-2`. This is a constant acceleration.