Question

A cheetah is resting when a hare traveling in a straight path passes at it's top speed of 25m/s. It takes the cheetah 1.5 seconds to get up and begin running. The cheetah accelerates to it's top speed of 45m/s is 20 seconds. It can maintain speed for 2 seconds before decelerating at 4 m/s^2.
A. What is the initial acceleration of the cheetah
B. How long does the chase last
C. How far has the cheetah traveled before stopping
D. What is the final velocity of the cheetah?

Answer 1

Answer:

a) 2.3 m/s^2

b) 33.3 s

c) 803 m

d) 0 m/s

Explanation:

We need to apply the accelerated motion formulas.

The acceleration is given by:

$a=\frac{v_f-v_o}{t}\\\\a=\frac{45m/s-0m/s}{20s}\\\\a=2.3m/s^2$

we need to calculate the time the cheetah takes to stop when it starts to decelerate:

$v=vo+a*t\\\\t=\frac{v-v_o}{a}\\\\t=\frac{0-45m/s}{-4m/s^2}=11.3s\\\\t_t=20s+2s+11.3s=33.3s$

We need to calculate the displacement for all of the stages:

stage 1:

$x_1=\frac{1}{2}*a*t^2\\\\x_1=\frac{1}{2}*2.3m/s^2*(20)^2\\\\x_1=460m$

stage 2:

$x_2=v*t\\x_2=45m/s*2s=90m$

stage 3:

$x_3=v_o*t+\frac{1}{2}*a*t^2\\\\x_3=45m/s*(11.3s)-\frac{1}{2}*4m/s^2*(11.3)^2\\\\x_3=253m$

The total displacement is given by:

$d=x_1+x_2+x_3\\d=460m+90m+253m=803m$

the final velocity of the cheetah is zero because the cheetah decelerates until stop.