A bus accelerates from rest at 1.5m/s^2 for 12s. It then travels for 25s after which it slows to a stop with an acceleration of magnitude 1.5m/s^2. (a) Find the total distance the bus travels And its average velocity of the trip.
x1 = (1/2)(1.5)(12^2) = 108 m
v1 = 1.5(12) = 18 m/s
x2 = 108 + 18(25) = 558 m
v2 = v1 = 18
x3 = 558 + 18 t - (1/2)1.5 t^2
v3 = 0
= 18 - 1.5 t
so t = 12 seconds to stop (of course)
then
x3 = 558 + 18(12) - (1/2)(1.5)(144)
= 558 + 216 - 108
= 666 meters
A bus accelerates from rest at 1.5m/s^2 for 12s. It then travels for 25s after which it slows to a stop with an acceleration of magnitude 1.5m/s^2. (a) Find the total distance the bus travel snd its average velocity of the trip.
666 meters
awdadsad
To solve this problem, we can break it down into three parts:
1. Finding the distance traveled during acceleration.
2. Finding the distance traveled during constant velocity.
3. Finding the distance traveled during deceleration.
1. Finding the distance traveled during acceleration:
The bus accelerates from rest at 1.5 m/s^2 for 12 seconds. To find the distance traveled during this time, we can use the formula:
s = ut + (1/2)at^2
where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.
Given that the initial velocity (u) is 0 m/s, the acceleration (a) is 1.5 m/s^2, and the time (t) is 12 seconds, we can substitute these values into the formula:
s = 0 * 12 + (1/2) * 1.5 * (12^2)
s = 0 + (1/2) * 1.5 * 144
s = 108 meters
Therefore, the distance traveled during acceleration is 108 meters.
2. Finding the distance traveled during constant velocity:
The bus travels at constant velocity for 25 seconds. Since the velocity is constant, the distance traveled is simply the product of the velocity and time:
d = v * t
where d is the distance, v is the velocity, and t is the time.
To find the velocity, we can use the formula for constant acceleration:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Since the bus is already at its final velocity after the acceleration phase, the initial velocity (u) is the same as the final velocity.
Given that the acceleration (a) is 1.5 m/s^2 and the time (t) is 12 seconds, we can substitute these values into the formula to find the velocity:
v = 0 + 1.5 * 12
v = 18 m/s
Now we can calculate the distance traveled during constant velocity:
d = 18 * 25
d = 450 meters
Therefore, the distance traveled during constant velocity is 450 meters.
3. Finding the distance traveled during deceleration:
The bus slows to a stop with an acceleration of magnitude 1.5 m/s^2. We can use the same formula as in step 1 to find the distance traveled during deceleration.
Given that the initial velocity is 18 m/s (from step 2), the acceleration is -1.5 m/s^2 (negative because it's deceleration), and the time is 25 seconds, we can substitute these values into the formula:
s = 18 * 25 + (1/2) * -1.5 * (25^2)
s = 450 - (1/2) * 1.5 * 625
s = 450 - 468.75
s = -18.75 meters
Since distance cannot be negative, we take the absolute value and get 18.75 meters.
Therefore, the distance traveled during deceleration is 18.75 meters.
Now, to find the total distance traveled, we can sum up the distances from all three parts:
Total distance = distance during acceleration + distance during constant velocity + distance during deceleration
Total distance = 108 + 450 + 18.75
Total distance = 576.75 meters
So, the total distance traveled by the bus is 576.75 meters.
To find the average velocity of the trip, we can use the formula:
average velocity = total distance / total time
Given that the total time is the sum of the times from all three parts: 12 + 25 + 25 = 62 seconds, we can substitute these values into the formula:
average velocity = 576.75 / 62
average velocity ≈ 9.3 m/s
Therefore, the average velocity of the trip is approximately 9.3 m/s.
a devil of a problem!
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