The cheetah can reach a top speed of 114 km/h (71 mi/h). While chasing its prey in a short sprint, a cheetah starts from rest an
d runs 45 m in a straight line, reaching a final speed of 72 km/h. (a) Determine the cheetah’s average acceleration during the short sprint, and (b) find its displacement at t 5 3.5 s.
2 answers:
Answer: a= 4.4ms-2
Displacement= 26.95 m
Explanation:
First, the speed in km/hr must be converted to m/s so that we can apply it in solving the question. The motion started from rest hence the initial velocity is 0m/s. The average displacement is also obtained from the equations of motion as shown in the image attached.
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A) Starting from rest, we have the entire cycle determined by 0.8s.
If we assume a constant movement, half of that time is when it reaches the highest point, that is, in 0.4s.
The distance as a function of speed and acceleration is given by,
At the initial point the speed is zero and the acceleration is equivalent to gravity.
B) When returning to the ground, the final speed is zero. Therefore, the equation that relates velocity to acceleration is given by,