Question

Cheetahs can accelerate to a speed of 20.4 m/s in 2.50 s and can continue to accelerate to reach a top speed of 29.0 m/s. direction the cheetah runs. Express the cheetah's top speed brop in miles per hour (mi/h). Assume the acceleration is constant until the top speed is reached and is zero thereafter. Let the +x direction point in tho mi/h Starting from a crouched position, how much time facci does it take a cheetah to reach its top speed and what distance d does it travel in that time? faceel If a cheetah sees a rabbit 126.0 m away, how much time feeal will it take to reach the rabbit, assuming the rabbit does not move and the cheetah starts from rest?

Answer 1

Solution:

vtop in miles per hour

=> v_top = 64.87 miles per hour

2) time taken by cheetah to reach 20.4 m/s is 2.5 seconds and the top speed is 29 m/s

Now using the equation

=> using the equation , v = u + at

=> a = 20.4 / 2.5

=> a = 8.16 m/s^2

Now again using the equation

=> v = u + at

=> 29 = 20.4 + 8.16 * t

=> t = 1.054 seconds

total time = 2.5 + 1.054

=> total time = 3.554 seconds

3) distance covered to reach 20.4 m/s

using the formula , s = u*t + 0.5 * a * t^2

=> s = 0 * t + 0.5 * 8.16 * 2.5 * 2.5

=> s = 25.5 meter

Now to reach to top speed distance covered

=> s1 = 20.4*1.054 + 0.5 * 8.16 * 1.054^2

=> s1 = 26.03 meter

total distance d = 51.5 meter

4) time it takes to reach top speed is 3.554 seconds

=> distance covered in this time = 0.5 * 8.16 * 3.554^2

=> d1 = 51.5 meter

Now the remaining distance will be covered with the constant speed which is 29.0 m/s

time taken = ( 126- 51.5) / 29

=> t1 = 2.57 seconds

total time = 2.57 + 2.5

=> t = 5.06 seconds (approx)

thanks.