A 6110-kg bus traveling at 20.0 m/s can be stopped in 24.0 s by gently applying the brakes. If the
driver slams on the brakes, the bus stops in 3.90 s. What is the average force exerted on the bus in both
these stops?
(Avg. force)*(time) = (momentum change)
= M * Vo
Solve for the average force.
hjbk
To find the average force exerted on the bus during both stops, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a):
F = m * a
In the first case, where the bus stops in 24.0 seconds, we can calculate the average acceleration using the formula:
a = (change in velocity) / (change in time)
The change in velocity is the initial velocity (20.0 m/s) because the bus stops completely, and the change in time is 24.0 seconds:
a = (0 - 20.0 m/s) / 24.0 s = -20.0 m/s / 24.0 s = -0.833 m/s^2
However, since we need to find the average force, we need to assume that the deceleration is constant throughout the stop. Therefore, we can take the absolute value of the average acceleration:
a = 0.833 m/s^2
Now, we can calculate the average force using the formula mentioned earlier:
F = m * a
F = 6110 kg * 0.833 m/s^2
F ≈ 5092.03 N
Therefore, the average force exerted on the bus in the first case is approximately 5092.03 Newtons.
In the second case, where the bus stops in 3.90 seconds, we can similarly calculate the average acceleration using the formula:
a = (change in velocity) / (change in time)
The change in velocity is again the initial velocity (20.0 m/s), and the change in time is 3.90 seconds:
a = (0 - 20.0 m/s) / 3.90 s = -20.0 m/s / 3.90 s = -5.13 m/s^2
Taking the absolute value of the average acceleration because the deceleration is assumed to be constant throughout the stop, we have:
a = 5.13 m/s^2
Now, we can calculate the average force using the formula mentioned earlier:
F = m * a
F = 6110 kg * 5.13 m/s^2
F ≈ 31366.23 N
Therefore, the average force exerted on the bus in the second case is approximately 31366.23 Newtons.
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