Cheetahs can accelerate to a speed of 20.0 m/s in 2.50 s and can continue to accelerate to reach a top speed of 30.0 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 (t-accel) does it take a cheetah to reach its top speed, and what distance (d) does it travel in that time?

The cheetah’s initial velocity is 0 m/s. Let’s use the following equation to determine the acceleration.

Explanation:

Acceleration = (v/t) = (20/2.5) = 8m/sec^2.

1) Time to 29m/sec. = (v/a) = (29/8) = 3.625 secs. (3.6)

Distance covered to top speed = (v^2/2a) = 52.5625 metres (52.6).

2) Distance to run at top speed = (120 - 52.5625) = 67.4375 metres.

Time = (d/v) = (67.4375/29) = 2.325 secs.

Total time to warthog = (2.325 + 3.625) = 5.95 secs.

i had this question and the answer was 3.15 miles.

martina travels from school to her mom's office. from there, she walks to school. after school, she walks to the library and then to a friends house before finally ending back at home. find martina's total distance traveled.

a) 0 miles  

b) 0.4 miles  

c) 1 mile  

d) 3.15 miles