An automobile starts from rest and travels down a straight section of road. The distance s (in feet) of the car from the starting position after t seconds is given by s (t) = 5t3 (a) Find the instantaneous velocity (in feet per second) at t=4 seconds.

Answer: 240 ft/sec

Explanation:

If the distance s, of the car regarding his starting position, is defined as a function of time, by definition, the instantaneous velocity at any time, is given by the derivative of the displacement (distance from origin) respect the time, as follows;

v = ds / dt ⇒ v = d/dt (5 t³) = 15 t² = 15 t²

Replacing by t=4 s, we get:

v = 15. 4² = 15.16 = 240 ft/sec