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:
[tex]a=\frac{v_f-v_o}{t}\\\\a=\frac{45m/s-0m/s}{20s}\\\\a=2.3m/s^2[/tex]
we need to calculate the time the cheetah takes to stop when it starts to decelerate:
[tex]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[/tex]
We need to calculate the displacement for all of the stages:
stage 1:
[tex]x_1=\frac{1}{2}*a*t^2\\\\x_1=\frac{1}{2}*2.3m/s^2*(20)^2\\\\x_1=460m[/tex]
stage 2:
[tex]x_2=v*t\\x_2=45m/s*2s=90m[/tex]
stage 3:
[tex]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[/tex]
The total displacement is given by:
[tex]d=x_1+x_2+x_3\\d=460m+90m+253m=803m[/tex]
the final velocity of the cheetah is zero because the cheetah decelerates until stop.
