A motorcyclist traveling 35.0 m/s realizes suddenly that he is headed straight for the edge of the Grand Canyon 50.0 m away. He immediately begins to brake, decreasing his speed at a rate of 7.00 m/s^2. With what velocity will he fly off into the canyon?
22.9 m/s
To find the velocity with which the motorcyclist will fly off into the canyon, we can use the equations of motion.
First, let's determine the time it takes for the motorcyclist to come to a stop. We can use the equation:
v = u + at
Where:
v = final velocity (0 m/s since the motorcyclist comes to a stop)
u = initial velocity (35.0 m/s)
a = acceleration (-7.00 m/s^2)
t = time (unknown)
Rearranging the equation to solve for time (t):
0 = 35.0 - 7.00t
7.00t = 35.0
t = 35.0 / 7.00
t ≈ 5.00 seconds
Now, with the time it takes to come to a stop, we can calculate the distance traveled during this time using the equation:
s = ut + (1/2)at^2
Where:
s = distance (unknown)
u = initial velocity (35.0 m/s)
t = time (5.00 seconds)
a = acceleration (-7.00 m/s^2)
Plugging in the values:
s = (35.0)(5.00) + (1/2)(-7.00)(5.00)^2
s = 175.0 - 87.5
s ≈ 87.5 meters
Now that we have the distance, we can find the velocity with which the motorcyclist will fly off into the canyon. Let's assume that the motorcyclist comes to a stop at some point before reaching the edge of the canyon. Then, the distance traveled during the deceleration is 87.5 meters.
The remaining distance to the canyon is given as 50.0 meters. Thus, we calculate the initial velocity before braking using the equation:
v^2 = u^2 + 2as
Where:
v = final velocity (unknown)
u = initial velocity (unknown)
a = acceleration (-7.00 m/s^2)
s = distance (50.0 meters)
Rearranging the equation to solve for initial velocity (u):
u^2 = v^2 - 2as
Since the final velocity is zero (stopped at the edge of the canyon):
u^2 = -2as
u = √(-2as)
u = √(-2)(-7.00)(87.5)
u ≈ 56.6 m/s
Therefore, the velocity with which the motorcyclist will fly off into the canyon is approximately 56.6 m/s.