A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by x(t)=bt2−ct3, where b = 3.00 m/s^2 and c = 0.100 m/s^3. Required:
a. Calculate the average velocity of the car for the time interval t = 0 to t = 8.0 s.
b. Calculate the instantaneous velocity of the car at the following times:

1. t = 0
2. t = 4.0 s
3. t = 8.0 s

c. How long after starting from rest is the car again at rest?

(a) average velocity = 17.6 m/s

(b) when t = 0, v = 0

    when t = 4, v = 19.2 m/s

    when t = 8, v = 28.8 m/s

(c) after starting from rest, the car will be at rest again in 20 s

Explanation:

Given;

x(t)=bt²−ct³, substitute the given values and the equation will become;

x(t)=3t²−0.1t³

(a)average velocity = total distance / total time

total distance, x(t) = 3t²−0.1t³

x(8) = 3t²−0.1t³

X(8) = 3(8)² - 0.1(8)³

X(8) = 140.8 m

total time = 8 s

average velocity = 140.8 / 8

average velocity = 17.6 m/s

(b) instantaneous velocity = dx / dt

dx / dt = 6t - 0.3t²

when t = 0

v = 0

when t = 4 s

v = 6(4) - 0.3(4²) = 19.2 m/s

when t = 8 s

v = 6(8) - 0.3(8²) = 28.8 m/s

(c) the velocity is zero at dx / dt = 0

6t - 0.3t² = 0

t(6 - 0.3t) = 0

t = 0    or  6 - 0.3t = 0

t = 0     or   0.3t = 6

t = 0      or   t = 6 / 0.3

t= 0       or    t =  20 s

After starting from rest, the car will be at rest again in 20 s

see earlier answer