When the shuttle bus comes to a sudden stop to avoid hitting a dog, it decelerates uniformly at 5.3 m/s2 as it slows from 9.5 m/s to 0 m/s.
Find the time interval of acceleration for the bus.
Answer in units of s
Vf=0
Vi=9.5m/s
A=5.3m/s2
T=?
Vf=Vi+At
0=9.5-5.3m/s2
-9.5=-5.3m/s2(t)
solve for T
T=1.8s
To find the time interval of acceleration for the bus, we can use the formula for acceleration:
acceleration = (final velocity - initial velocity) / time interval
Given:
Initial velocity (u) = 9.5 m/s
Final velocity (v) = 0 m/s
Acceleration (a) = -5.3 m/s^2 (negative because it is decelerating)
Plugging these values into the formula, we have:
-5.3 m/s^2 = (0 m/s - 9.5 m/s) / time interval
Now we can solve for the time interval:
-5.3 m/s^2 * time interval = -9.5 m/s
Dividing both sides of the equation by -5.3 m/s^2:
time interval = -9.5 m/s / -5.3 m/s^2
Simplifying:
time interval = 1.792452830188679 s
Therefore, the time interval of acceleration for the bus is approximately 1.792 s.
To find the time interval of acceleration for the bus, we need to use the equation of motion:
v = u + at
where:
v = final velocity (0 m/s)
u = initial velocity (9.5 m/s)
a = acceleration (-5.3 m/s²)
t = time interval
Rearranging the equation to solve for time (t), we have:
t = (v - u) / a
Substituting the given values into the equation, we get:
t = (0 - 9.5) / (-5.3)
Calculating this expression, we find:
t ≈ 1.7925 seconds
Therefore, the time interval of acceleration for the bus is approximately 1.7925 seconds.