The cheetah can reach a top speed of 114km/h (71mi/h). While chasing its prey in a short sprint, a cheetah starts from rest and runs 50m in a straight line reaching a final speed of 91 km/h.
a.) Determine the cheetah's average acceleration during the short sprint.
b.) Find it's displacement at t=3.0s. (assume the cheetah maintains a constant acceleration throughout the sprint)
Speed of Cheetah is 20m/s
change in speed = 91 km/hr = 91,000 m /3600 s
= 25.3 m/s
average speed if a is constant = 12.6 m/s
time for 50 m = 50/12.6 = 3.96 seconds (wow!)
average a = change in speed / time
= 25.3 / 3.96 = 6.4 m/s^2
at 3 seconds
v = a t = 6.4 * 3 = 19.2 m/s
average v for 3 seconds = 19.2/2 = 9.59 m/s
so
distance = 9.59 * 3 = 28.8 meters
To solve these problems, we need to use the equations of motion, specifically the equation for average acceleration and the equation for displacement.
a.) Determine the cheetah's average acceleration during the short sprint.
The equation for average acceleration is:
average acceleration = (final velocity - initial velocity) / time
Given:
Initial velocity (u) = 0 km/h
Final velocity (v) = 91 km/h
Time (t) = Unknown
We need to solve for time.
Using the given information:
91 km/h = (0 km/h + 114 km/h) / 2
182 km/h = 114 km/h * 2
182 km/h = 228 km/h
t = time = 1 hour
Convert time to seconds:
1 hour = 3600 seconds
Substituting the values:
average acceleration = (91 km/h - 0 km/h) / 3600s
average acceleration = 91 km/h / 3600s
average acceleration = 0.02528 km/h/s
Therefore, the cheetah's average acceleration during the short sprint is approximately 0.02528 km/h/s.
b.) Find its displacement at t=3.0s. (assume the cheetah maintains a constant acceleration throughout the sprint)
The equation for displacement is:
displacement (s) = initial velocity (u) * time (t) + (1/2) * acceleration (a) * (time^2)
Given:
Initial velocity (u) = 0 km/h
Time (t) = 3.0s
Acceleration (a) = 0.02528 km/h/s
Substituting the values:
displacement (s) = 0 km/h * 3.0s + (1/2) * 0.02528 km/h/s * (3.0s)^2
displacement (s) = 0 + (1/2) * 0.02528 km/h/s * 9.0s^2
displacement (s) = 0 + 0.11376 km/h * 9.0s
displacement (s) = 1.02384 km/h * 9.0s
displacement (s) = 9.21456 km * h's
displacement (s) = 9214.56 m
Therefore, at t = 3.0s, the cheetah's displacement is approximately 9214.56 meters.
To determine the cheetah's average acceleration during the short sprint, we can use the formula:
acceleration = (final velocity - initial velocity) / time
First, we need to convert the velocities from km/h to m/s, since the acceleration formula requires m/s units.
1 km/h = 1000 m/3600 s = 5/18 m/s
The final velocity is 91 km/h, so in m/s it will be:
91 km/h * (5/18 m/s per 1 km/h) = 25.28 m/s
The initial velocity is 0 m/s since the cheetah starts from rest.
The time is not given, but we know the cheetah runs 50m, so we can calculate the time using the equation of motion for constant acceleration:
displacement = initial velocity * time + (1/2) * acceleration * time^2
Since the initial velocity is 0 m/s, the equation simplifies to:
displacement = (1/2) * acceleration * time^2
Rearranging the equation, we can solve for time:
time = sqrt(2 * displacement / acceleration)
Plugging in the values, we get:
time = sqrt(2 * 50m / acceleration)
Now we can substitute the time value back into the acceleration formula to calculate the average acceleration.
a.) Average acceleration:
acceleration = (final velocity - initial velocity) / time
acceleration = (25.28 m/s - 0 m/s) / time
b.) Displacement at t=3.0s:
time = 3.0s
displacement = (1/2) * acceleration * time^2
Now we just need to substitute the values and calculate the answers.