A truck on a straight road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 11.0 m/s. Then the truck travels for 65.0 s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.00 s.
(a) How long is the truck in motion?
s
(b) What is the average velocity of the truck for the motion described?
To find the answers to these questions, we need to break down the truck's motion into three parts: acceleration, constant speed, and deceleration.
(a) How long is the truck in motion?
1. First, we need to calculate the time it takes for the truck to reach its maximum speed of 11.0 m/s. We can use the equation:
vf = vi + at
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.
Rearranging the equation, we get:
t = (vf - vi) / a
Substituting the given values:
t = (11.0 m/s - 0 m/s) / 2.00 m/s²
t = 5.5 s
So, the time taken to accelerate to the maximum speed is 5.5 seconds.
2. Next, we need to find the distance traveled during this acceleration phase. We can use the equation:
d = vit + 0.5at²
where d is the distance, vi is the initial velocity, a is the acceleration, and t is the time.
Substituting the known values:
d = 0 m/s * 5.5 s + 0.5 * 2.00 m/s² * (5.5 s)²
d = 0 + 0.5 * 2.00 m/s² * 30.25 s²
d = 30.25 m
So, the distance traveled during the acceleration phase is 30.25 meters.
3. After reaching the maximum speed, the truck travels at a constant velocity for 65.0 seconds.
4. Lastly, the truck decelerates uniformly to a stop in an additional 5.00 seconds.
To find the total time the truck is in motion, we add the times taken for each phase:
Total time = accelerating time + constant speed time + decelerating time
= 5.5 s + 65.0 s + 5.00 s
= 75.5 s
So, the truck is in motion for 75.5 seconds.
(b) What is the average velocity of the truck for the motion described?
Average velocity is calculated by dividing the total distance traveled by the total time taken.
1. The distance traveled during acceleration is 30.25 meters.
2. The distance traveled during the constant speed phase is given by the formula:
d = vt
where d is the distance, v is the velocity, and t is the time.
Since the speed is constant at 11.0 m/s and the time is 65.0 seconds:
d = 11.0 m/s * 65.0 s
d = 715.0 m
So, the distance traveled during the constant speed phase is 715.0 meters.
3. The distance traveled during deceleration can be found using the same formula:
d = vt
Since the final velocity is 0.0 m/s and the time is 5.00 seconds:
d = 0.0 m/s * 5.00 s
d = 0.0 m
So, the distance traveled during deceleration is 0.0 meters (since the truck comes to a stop).
The total distance traveled is the sum of the distances from each phase:
Total distance = distance during acceleration + distance during constant speed + distance during deceleration
= 30.25 m + 715.0 m + 0.0 m
= 745.25 m
The average velocity is then:
Average velocity = Total distance / Total time
= 745.25 m / 75.5 s
= 9.84 m/s
So, the average velocity of the truck for the described motion is 9.84 m/s.