This is a trig question that is too long to put in the title. See description

Question

A small pebble stuck on the tread of a car’s tire is travelling at a linear

speed of 35 miles per hour. If the radius of the tire were 1 inch greater and its

angular velocity were the same, the pebble would be travelling 5 miles per hour

faster. How high above the ground is the pebble a tenth of a second after it touches

the ground?

Answer 1

I would hope this is for extra credit...

Let's let A = rotations of the tire per hour.

Then: A * 2 * pi * r = distance traveled in an hour where r is radius of the tire

A * 2 * pi * r = 35 * 5280 *12

A * 2 * pi * (r + 1) = 40 * 5280 * 12 (if 1" greater, it would travel 5 mph faster)

The previous two imply:

A * 2 * pi = 5 * 5280 * 12

A = 50420.29 revolutions per hour. (or 840.3382 rpm)

Now that we have A, we can calculate r

A * 2 * pi * r = 35 * 5280 *12

50420.29 * 2 * pi * r = 35 * 5280 *12

r = 35 * 5280 * 12 / (50420.29 * 2 * pi)

r = 7 inches

So now, after 1/10 of a second, the wheel as made 1/10 of 1/3600 of A rotations:

1/10 * 1/3600 * 50420.29 = 1.400564 rotations in 1/10 of a second

So the wheel has made 1.4 rotations. To compute the height of the pebble, which isn't as simple as you might think, we need to start with it at ground level (z= 0). The wheel's axle is a z = 7".

7 - 7 * cos(1.4 * 2 * pi) = 12.66312 inches above the ground.

Drops mic, stumbles off wondering why anyone would ask this question....