How many feet does the car travel before coming to a complete stop?

Question

A car traveling at 40 ft/sec decelerates at a constant 5 feet per second squared. How many feet does the car travel before coming to a complete stop?

Answer 1

Let a be the acceleration function,

v be the velocity function, and

s be the position function of the car.

"A car ... decelerates at a constant 5 feet per second squared"

so a(t) = -5

v(t) = ∫ a(t) dt = ∫ -5 dt = -5t + C

"A car traveling at 40 ft/sec"

so v(0) = 40

v(0) = 40 ⇒ -5 * 0 + C = 40 ⇒ C = 40

Thus

v(t) = -5t + 40

When does the car come to a complete stop?

The car comes to a complete stop at time t* when v(t*) = 0.

v(t*) = 0

⇒ -5t* + 40 = 0

⇒ -5t* + 40 - 40 = 0 - 40

⇒ -5t* = -40

⇒ -5t* / (-5) = -40 / (-5)

⇒ t* = 8

How many feet does the car travel before coming to a complete stop?

This question is answered by integrating the velocity function between time 0 and the stopping time 8.

∫₀⁸ v(t) dt

= ∫₀⁸ (-5t + 40) dt

= (-5/2)t² + 40t |₀⁸

= (-5/2) * 8² + 40 * 8 - [(-5/2)* 0² + 40 * 0]

= (-5/2) * 64 + 320

= -5 * 32 + 320

= -160 + 320

= 160