Question

Cheetahs can accelerate to a speed of 21.8 m/s in 2.55 s and can continue to accelerate to reach a top speed of 28.1 m/s . Assume the acceleration is constant until the top speed is reached and is zero thereafter. Let the +x direction point in the direction the cheetah runs. Express the cheetah's top speed vtop in miles per hour (mi/h) .

Starting from a crouched position, how much time totall does it take a cheetah to reach its top speed and what distance d does it travel in that time?
If a cheetah sees a rabbit 122.0 m away, how much time ttotal will it take to reach the rabbit, assuming the rabbit does not move and the cheetah starts from rest?

Answer 1

Step-by-step explanation:

Given

Cheetah speed=21.8 m/s in 2.55 sec

i.e. its
`a=(21.8)/(2.55)=8.55 m/s^2`

Cheetah top speed=28.1 m/s

And 1 m/s is equal to 2.23694 mph

therefore 28.1 m/s is
`2.236* 28.1=62.858 mph`

Time taken to reach top speed

v=u+at

`28.1=0+8.55* t`

`t=(28.1)/(8.55)=3.286 s`

Distance traveled during this time

`v^2-u^2=2as`

`28.1^2=2* 8.55* s`

`s=(789.61)/(2* 8.55)=46.176 m`

If cheetah sees a rabbit 122 m away

time taken to reach rabbit

`s=ut+(1)/(2)at^2`

`122=0+(1)/(2)* 8.55* t^2`

`t^2=(244)/(8.55)`

`t=√(28.53)=5.34 s`