Question

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 and 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.

Answer 1

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.

Answer 2

Answer:

a) Average acceleration =4.44m/s^2

b)6354.20m

Explanation:

Change in speed= 72000/3600= 20m/s

Average speed= 20m/s/2=10m/s

Time to run 45m= 45/10 =4.5seconds

Average acceleration = change in speed/ time= 20/4.5=4.44m/s^2

b) V =at at t=53.5second

V = 4.44×53.5=237.54m/s

Average speed= 237.54/2=218.77m/s

Displacement= average speed×time= 118.77× 53.5

Displacement= 6354.20m