# Acceleration, speed, distance problem

1. Feb 4, 2004

### Raffi

A motorcycle enters the freeway 8 seconds before a car following it. The motorcycle upon entering the freeway accelerates to 68 mph in 3 seconds. How fast will the car have to travel in order to catch up with the motorcycle in 1 mile. Keep in mind, the car does 0-60 in 9 seconds. If possible, please show me the formula for solving, as I'm sure the Judge may want to see it.
Let me know if you need anymore numbers.
Thank you

2. Feb 4, 2004

### Raffi

I've tried to solve this problem numerous times but am stumped by the fact that 2 vehicles are invloved and can't seem to figure out the formula for the car's speed.

3. Feb 5, 2004

### Xamfy19

Upon entering freeway, do you assume car and motorcycle is accelerating from 0? or they have initial speed?

4. Feb 18, 2004

### Raffi

problem solving help

Both are starting from zero.

5. Feb 18, 2004

### Xamfy19

I got 91.36 mph to catch motorcycle.
If this is reasonable, then I will show my work later.

6. Feb 18, 2004

### Raffi

Thanks, I'd really appreciate it if you could let me know how you came to that figure...

7. Feb 18, 2004

### Xamfy19

formulas

a1 = (68*5280/3600)/3 = 33.244 ft/s^2
Now, calculate how far within the first 3 seconds the motorcycle traveled.
d1 = 0.5*33.244*9 = 149.6 ft
Calculate the remaining distance before 1 mile = 5280 - 149.6 = 5130.4 ft.
Time to travel 5130.4 ft = 5130.4/99.73 = 51.44 seconds
Total motorcycle traveled time = 51.44 + 3 = 54.44 seconds

Since the car entered hwy 8 seconds later, that left 46.44 seconds for the car to catch up the motorcycle.

From the given information, we have to assume the car accelerated from 0 to a speed at constant acceleration, otherwise there is no solution (or many solutions) to this question.

Let's calculate the acceleration of the car.
a2= 60*5280/3600*9 = 9.77 ft/s^2
which was slower than motorcycle. However, the car keep on accelerating up to a speed. Let's find out the acceleration time (during which the car accelerated)
Let d2 = the distance when the car accelerated,
V = the ultimate speed when the car cruising
t = the time the car accelerated

d2 = 0.5 * 9.77 * t^2
V = 9.77 * t
(5250 - d) / (46.44 - t) = V,

Solve these three equations, you will get t = 13.7 seconds (anothe answer, 79 seconds wasn't right)

V = 9.77 * 13.7 = 133.8 ft/s
= 91.2 mph.

That's how I got this answer.