Physics velocity/acceleration question...?

Question

A car is travelling at 60 MPH when its brakes are fully applied, creating a constant deceleration of 22 ft/sec. Find the distance covered by the car before it comes to a stop. (Note 60 MPH=88 ft/sec.) I want to just say that it will take 4 seconds to stop(since 22x4 is 88) and then plug that into the distance=vt formula, but I know this isn't right. I am supposed to use antidifferentiation in some way. The question is multiple choice, so here are the choices:

A)172 ft
B)176 ft
C)174 ft
D)180 ft
E)178 ft

Answer 1

First, the units for deceleration (same as for acceleration) are "distance/time^2," not "distance/time" which would be a velocity.

First, the formula way (without using integration, i.e., antidifferentiation):

Assuming you meant to say 22 ft/sec^2, we can get the time to stop (assuming constant deceleration as you have said):

t is the time
Vf is the "final" velocity
Vi is the "initial" velocity
d is the distance

t = (Vf-Vi)/a
t = (0-88)/-22 = -88/-22
t = 4

Plug the time of 4 seconds into the formula:

d = 1/2 * ( Vf + Vi ) * t
d = 1/2 * ( 0 + 88) * 4
d = 176

To solve this thing using antidifferentiation, you need to use integrals. I'll leave the solution for time (4 secs is the answer) to you. Then the distance (s below; the capital S is the integral symbol):

s(t) = S v(t) dt
s(t) = S 22t dt
s(t) = 1/2*22*t^2
s(t) = 1/2*22*4^2 = .5*22*16
= 176

Answer 2

I'm sure it was just a typo when you said "deceleration of 22 ft/sec"??? The units of acceleration or deceleration must be distance/time^2. So I'll correct what you said and use a deceleration of 22 ft/sec^2.

I may not work this the way you need. Antidifferentiation? I don't even know that term, perhaps it's what I know as integration? Anyway, my way doesn't require calculus.

The way you started to work it would have worked if you remembered that you need average velocity over the 4 seconds. It's 88 ft/sec at the beginning and 0 at the end. Average = 44 ft/sec. Then times t would give you 176 ft.

You probably have several formulas that you were given for this type problem. I choose a different one for this problem and just the one formula does the job.

V^2 = Vo^2 + 2as
0^2 = (88 ft/sec)^2 + 2*(-22 ft/sec^2)*s
s = (88^2 ft^2/sec^2) / 44 ft/sec^2 = 7744 / 44 ft = 176 ft

Answer 3

Make a graph where the vertical is the speed, the horizontal is the seconds. Plot the speed at every point (this is easy since the change in acceleration is constant) and then measure the area under the graph.