Answer :
Missing parts in the text of the exercise:
- The distance from the traffic light is given by [tex]x(t) = bt^2 -ct^3[/tex], where [tex]b=2.4 m/s^2[/tex] and [tex]c=0.13 m/s^3[/tex]
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:
[tex]v_{ave} = \frac{x(8.0 s)-x(0 s)}{8.0 s-0 s} [/tex]
x(0 s), the distance covered after t=0 s, is zero, while the distance after t=8.0 s is
[tex]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=[/tex]
[tex]=87.04 m[/tex]
Therefore, the average velocity is
[tex]v_{ave}= \frac{x(8.0 s)-x(0)}{8.0s-0}= \frac{87.04 m}{8.0s}= 10.88 m/s [/tex]
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):
[tex]v(t) = \frac{dx(t)}{dt}=2bt-3ct^2 [/tex]
So now we just have to substitute t=0, t=4 s and t=8 s:
- t=0: v(0)=0
- t=4 s: [tex]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[/tex]
- t=8 s: [tex]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[/tex]
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:
[tex]2bt-3ct^2=0[/tex]
[tex]t(2b-3ct)=0[/tex]
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
[tex]t= \frac{2b}{3c}= \frac{2\cdot 2.4 m/s^2}{3 \cdot 0.13 m/s^3}=12.31 s [/tex]
and this is the time at which the car is at rest again.
- The distance from the traffic light is given by [tex]x(t) = bt^2 -ct^3[/tex], where [tex]b=2.4 m/s^2[/tex] and [tex]c=0.13 m/s^3[/tex]
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:
[tex]v_{ave} = \frac{x(8.0 s)-x(0 s)}{8.0 s-0 s} [/tex]
x(0 s), the distance covered after t=0 s, is zero, while the distance after t=8.0 s is
[tex]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=[/tex]
[tex]=87.04 m[/tex]
Therefore, the average velocity is
[tex]v_{ave}= \frac{x(8.0 s)-x(0)}{8.0s-0}= \frac{87.04 m}{8.0s}= 10.88 m/s [/tex]
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):
[tex]v(t) = \frac{dx(t)}{dt}=2bt-3ct^2 [/tex]
So now we just have to substitute t=0, t=4 s and t=8 s:
- t=0: v(0)=0
- t=4 s: [tex]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[/tex]
- t=8 s: [tex]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[/tex]
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:
[tex]2bt-3ct^2=0[/tex]
[tex]t(2b-3ct)=0[/tex]
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
[tex]t= \frac{2b}{3c}= \frac{2\cdot 2.4 m/s^2}{3 \cdot 0.13 m/s^3}=12.31 s [/tex]
and this is the time at which the car is at rest again.