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

answer: option c. i.e. electric current flowing “downhill” from the negative electrode to the positive electrode in a battery

explanation:

the water analogy to the circuits define the pipes (water flows) to the electric current. now, we know that the electric current flows from the positive terminal to the circuit (as shown in the diagram attached).

hence, option c is the correct answer.