Question

12*3a car is stopped at a traffic light. it then travels along a straight road so that its distance from the light is given by where and (a) calculate the average velocity of the car for the time interval to (b) calculate the instantaneous velocity of the car at and (c) how long after starting from rest is the car again at rest?

Answer 1

Missing parts in the text of the exercise:
- The distance from the traffic light is given by $x(t) = bt^2 -ct^3$, where $b=2.4 m/s^2$ and $c=0.13 m/s^3$

Solution:

part a) calculate the average velocity of the car for the time interval t=0 to t=8.0 s
- The average velocity is given by the ratio between the distance covered in the time interval:
$v_{ave} = \frac{x(8.0 s)-x(0 s)}{8.0 s-0 s}$
x(0 s), the distance covered after t=0 s, is zero, while the distance after t=8.0 s is
$x(8.0 s)=b(8 s)^2-c(8 s)^3 = (2.4 m/s^2)(8 s)^2 -(0.13 m/s^3)(8 s)^3=$
$=87.04 m$
Therefore, the average velocity is
$v_{ave}= \frac{x(8.0 s)-x(0)}{8.0s-0}= \frac{87.04 m}{8.0s}= 10.88 m/s$

part b) calculate the instantaneous velocity of the car at t=0, t=4.0 s and t=8.0 s
- The instantaneous velocity can be found by performing the derivation of x(t):
$v(t) = \frac{dx(t)}{dt}=2bt-3ct^2$
So now we just have to substitute t=0, t=4 s and t=8 s:
- t=0: v(0)=0
- t=4 s: $v(4.0 s)=2(2.4 m/s^2)(4.0 s)-3(0.13 m/s^2)(4.0s)^2=12.96 m/s$
- t=8 s: $v(8.0 s)=2(2.4 m/s^2)(8.0 s)-3(0.13 m/s^2)(8.0s)^2=13.44 m/s$

part c) how long after starting from rest is the car again at rest?
- To solve this part we must find the value of t for which v(t)=0, so:
$2bt-3ct^2=0$
$t(2b-3ct)=0$
The first solution is t=0 s, which corresponds to the beginning of the motion, so we are not interested in this value. The second solution is
$t= \frac{2b}{3c}= \frac{2\cdot 2.4 m/s^2}{3 \cdot 0.13 m/s^3}=12.31 s$
and this is the time at which the car is at rest again.