5. A race car travels on a racetrack at 44 m/s and slows at a con- stant rate to a velocity of 22 m/sover 11 s. How far does it move during this time?
6. A truck is traveling at 22 m/s when the driver notices a speed limit sign for the town ahead. He slows down to a speed of 14 m/s. He travels a distance of 125 m while he is slowing down. (a) Calculate the acceleration of the truck. (b) How long did it take the truck driver to change his speed?
For question 5:
We are given:
Initial velocity (u) = 44 m/s
Final velocity (v) = 22 m/s
Time (t) = 11 s
Using the formula for average velocity:
Average velocity = (u + v)/2
Average velocity = (44 m/s + 22 m/s)/2
Average velocity = 66 m/s/2
Average velocity = 33 m/s
Using the formula for displacement:
Displacement = Average velocity * time
Displacement = 33 m/s * 11 s
Displacement = 363 m
Therefore, the race car moves a distance of 363 meters.
For question 6:
We are given:
Initial velocity (u) = 22 m/s
Final velocity (v) = 14 m/s
Distance (s) = 125 m
Using the formula for acceleration:
Acceleration (a) = (v - u)/t
Substituting the given values:
Acceleration = (14 m/s - 22 m/s)/t
Acceleration = -8 m/s / t
To find the time (t), we can rearrange the equation:
t = -8 m/s / acceleration
t = -8 m/s / (-8 m/s^2)
t = 1 s
Therefore, it took the truck driver 1 second to change his speed.
Please note that for question 6, the truck driver was given the distance while he was slowing down, not the time. To calculate the acceleration, we need the time, so it is assumed that the truck driver traveled the given distance in a straight line with a constant acceleration.