Answer:
(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
