on turning a corner a motorcyclist rushing
at 44 m/s. finds a child on the road 100 m away. he applies the brake so as he stops the motorcycle within 1m of the child. calculate the time required to stop?
vf^2=vi^2+2ad
d=99m
solve for a
then,
vf=vi+at solve for t.
2.25 seconds
To calculate the time required to stop the motorcycle, we can use kinematic equations. We'll use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, as the motorcycle stops)
u = initial velocity (44 m/s)
a = acceleration (unknown, to be determined)
s = distance (100 m)
First, let's rearrange the equation to solve for acceleration:
a = (v^2 - u^2) / (2s)
Substituting the given values:
a = (0^2 - 44^2) / (2 * 100)
Simplifying:
a = (-44)^2 / 200
a = 1936 / 200
a = 9.68 m/s^2
Now that we have the acceleration, we can use another kinematic equation to find the time required to stop:
v = u + at
Where:
v = final velocity (0 m/s)
u = initial velocity (44 m/s)
a = acceleration (-9.68 m/s^2, negative because it's in the opposite direction of motion)
t = time (unknown, to be determined)
Rearranging the equation to solve for time:
t = (v - u) / a
Substituting the given values:
t = (0 - 44) / -9.68
Simplifying:
t = 44 / 9.68
t ≈ 4.55 seconds
Therefore, it would take approximately 4.55 seconds for the motorcyclist to stop the motorcycle and reach the child within 1 meter.