A car is traveling at a rate of 18 feet per second when the brakes are applied.

Question

A car is traveling at a rate of 18 feet per second when the brakes are applied. The position function for the car is given by s = –2.25t2 + 18t, where s is measured in feet and t is measured in seconds. Create a table showing the position, velocity, and acceleration for each given value of t.
t 0 1 2 3 4 5

s(t) _____ 15.75 27 _____ ______ 33.75

v(t) _____ 13.50 9 4.50 ______ ______

a(t) -4.50 -4.50 -4.50 -4.50 -4.50 -4.50


What can you conclude?
A) It takes 4 seconds for the car to stop, at which time it has traveled 4.5 ft.
B) It takes 3 seconds for the car to stop, at which time it has traveled 33.75 ft.
C) It takes 4 seconds for the car to stop, at which time it has traveled 36 ft.
D) It takes 4 seconds for the car to stop, at which time it has traveled 2.25 ft.
E) It takes 5 seconds for the car to stop, at which time it has traveled 33.75 ft.
F) It takes 4 seconds for the car to stop, at which time it has traveled 18 ft.

What equation did you use for v(t)? that is where I got stuck.

Explanation / Answer

s = –2.25t^2 + 18t remember: s=1/2*a*t^2+v0t. So a has to be 2 * -2.25 and v0 has to be 18. v=a*t+v0 v= –4.5t + 18 and v to a: a=-4.5 just plug in. So like for s(5)=-2.25*5^2+18*5=33.75 s(t) 0 15.75 27 33.75 36 33.75 v(t) 18 13.50 9 4.50 0 -4.5 a(t) -4.50 -4.50 -4.50 -4.50 -4.50 -4.50 C) It takes 4 seconds for the car to stop, at which time it has traveled 36 ft.

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