When the shuttle bus comes to a sudden stop
to avoid hitting a dog, it decelerates uniformly
at 3.6 m/s^2
as it slows from 8.5 m/s to 0 m/s.
Find the time interval of acceleration for
the bus.
Answer in units of s.
(8.5 m/s) / (3.6 m/s^2) = 2.36 s
To find the time interval of acceleration for the bus, we can use the equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity = 0 m/s
u = initial velocity = 8.5 m/s
a = acceleration = -3.6 m/s^2 (negative because the bus is decelerating)
s = displacement (unknown)
Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)
Plugging in the given values, we get:
s = (0^2 - 8.5^2) / (2*(-3.6))
s = (-72.25) / (-7.2)
s = 10.0347 m
The displacement during the deceleration is approximately 10.0347 m.
Now, we can use another equation of motion to find the time interval:
v = u + at
where:
v = final velocity = 0 m/s
u = initial velocity = 8.5 m/s
a = acceleration = -3.6 m/s^2 (negative because the bus is decelerating)
t = time (unknown)
Plugging in the given values, we get:
0 = 8.5 + (-3.6) * t
Simplifying the equation:
-3.6t = -8.5
Dividing both sides by -3.6:
t = (-8.5) / (-3.6)
t ≈ 2.36 s
Therefore, the time interval of acceleration for the bus is approximately 2.36 seconds.
To find the time interval of acceleration for the bus, we need to use the equation of motion that relates velocity, acceleration, and time. The equation is:
v = u + at
where:
v = final velocity (0 m/s)
u = initial velocity (8.5 m/s)
a = acceleration (-3.6 m/s^2)
t = time
We need to solve for time (t) in this equation. Rearranging the equation, we get:
t = (v - u) / a
Substituting the given values into the equation:
t = (0 - 8.5) / (-3.6)
Now, let's calculate:
t = (-8.5) / (-3.6)
t ≈ 2.36s
Therefore, the time interval of acceleration for the bus is approximately 2.36 seconds.