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?

1 Answer
Jan 21, 2018

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