Answer
(a) The average velocity is 12.0 m/s
(b) At t = 0, v = 0 m/s
At t = 5.0 s, v = 15.0 m/s
At t = 10.0 s, v = 12.0 m/s
(c) The car is at rest again after t = 13.3 seconds.
Work Step by Step
$x(t) = b t^2 - c t^3$
$b = 2.40~m/s^2$
$c = 0.120~m/s^3$
(a) At t = 0,
$x = (2.40~m/s^2)(0)^2 - (0.120~m/s^3)(0)^3 = 0$
At t = 10.0 s,
$x = (2.40~m/s^2)(10.0~s)^2 - (0.120~m/s^3)(10.0~s)^3$
$x = 240~m - 120~m = 120~m$
$average~velocity = \frac{\Delta x}{\Delta t} = \frac{120~m- 0}{10.0~s} = 12.0~m/s$
(b) $x(t) = b t^2 - c t^3$
$v(t) = \frac{dx}{dt} = 2bt - 3ct^2$
$b = 2.40~m/s^2$
$c = 0.120~m/s^3$
At t = 0,
$v = (2)(2.40~m/s^2)(0) - (3)(0.120~m/s^3)(0)^2 = 0$
At t = 5.0 s,
$v = (2)(2.40~m/s^2)(5.0~s) - (3)(0.120~m/s^3)(5.0~s)^2$
$v = 24.0~m/s - 9.0~m/s = 15.0~m/s$
At t = 10.0 s,
$v = (2)(2.40~m/s^2)(10.0~s) - (3)(0.120~m/s^3)(10.0~s)^2$
$v = 48.0~m/s - 36.0~m/s = 12.0~m/s$
(c) The car is at rest when v = 0.
$v = (2)(2.40~m/s^2)(t) - (3)(0.120~m/s^3)(t)^2 = 0$
$(t)((4.80~m/s^2) - (0.360~m/s^3)(t)) = 0$
$t = 0$ or $(4.80~m/s^2) - (0.360~m/s^3)(t) = 0$
$(4.80~m/s^2) = (0.360~m/s^3)(t)$
$t = \frac{4.80~m/s^2}{0.360~m/s^3} = 13.3~s$
The car is at rest again after t = 13.3 seconds.
