If a cheetah sees a rabbit 132.0 m away, how much time t total will it take to reach the rabbit, assuming the rabbit does not move and the cheetah starts from rest?

Question

Cheetahs can accelerate to a speed of 20.0 m/s in 2.50s  and can continue to accelerate to reach a top speed of 29.5 m/s Assume the acceleration is constant until the top speed is reached and is zero thereafter.

taceel= 3.69s (the time it takes to reach its top speed from crouching position)

d=54.39m (distance it travels in that time)

I solve those two correctly but I got this particular question wrong

I did 132m =0 + 1/2 * 8.00 *t^2 then 264/8.00 and square rooted the result and got 5.74s but it said it was wrong because it was too small a value and the problem's hint was that the cheetah only accelerates over the shorter distance needed to reach its top speed. Upon attaining its top speed, the cheetah continues at that constant speed until it catches the rabbit.

Answer 1

Hello Javier,

You are using the correct basic equation to answer this question.

d = d₀ + v₀t + 1/2 at²

But, we have different values for the constants when the cheetah is accelerating and when it reaches its top speed.

During the acceleration

d₀ = 0 m

v₀ = 0 m/s

a = 8 m/s² calculated below

Once the cheetah reaches top speed of 29.5 m/s the acceleration goes to 0 m/s²

So after the cheetah reaches top speed the constants are

d₀ = 54.39 m calculated below

v₀ = 29.5 m/s

a₀= 0 m/s

First we will calculate the acceleration. Then with this acceleration, we can calculate the time to reach the top speed and the distance traveled to reach the top speed of 29.5 m/s.

Then we will determine the time to travel the remaining distance to reach 132 meters traveling at a constant speed of 29.5 m/s.

The total time will then be the time to reach the top speed of 29.5 m/s plus the time traveling at 29.5 m/s to cover the entire 132 m

First lets calculate the acceleration

a=(v-v₀)/t = 20 m/s/2.5 s= 8 m/s²

Then lets calculate the time to reach the top speed of 29.5 m/s

t = 29.5 m/s/8 m/s² = 3.6875 s

The distance traveled while accelerating is given by

d = d₀ + v₀t + 1/2 at²

d = 0 m + (0 m/s)(3.6875 s) + 1/2(8 m/s²)(3.6875 s)² = 54.39 m

The remaining distance is covered by the cheetah at a constant speed of 29.5 m/s

d = d_{0 +} v t

132 m = 54.39 m + (29.5 m/s)(t)

Solving for t we get

t = (132 m - 54.39 m)/(29.5 m/s) = 2.63 s

Now we add the acceleration time and the time traveling at a constant speed to get the total time.

t = 3.69 s + 2.63 s = 6.32 s

It looks like you have the basic idea.

What you calculated is the time it would take if the cheetah continued to accelerate at 8 m/s² until it reached the rabbit.

Hope this helps.

Please let me know if you have any questions.