Question

An object moves along a straight line. Its displacement s (in metres) from a reference point at time t (in seconds) is given by s(t) = 4t^3 - 21t^2 + 18t + 5 (t ≥ 0) expressions for the velocity v(t) and the acceleration a(t) of the object at time t?

Answer 1

I tried this:

Explanation:

Here we can use the idea that:
velocity is the change of position with time;
acceleration is the change of velocity with time.

But, change in mathematics is represented by the DERIVATIVE.
So we can derive our expression with respect to time `t` to find velocity `v` and acceleration `a`.

Given:
`s(t)=4t^3-21t^2+18t+5` in `"meters"`

we derive:
`v(t)=(ds(t))/dt=12t^2-42t+18` in `"meters"/"seconds"`
again:
`a(t)=(dv(t))/dt=24t-42` in `"meters"/"seconds"^2`