# Homework Help: Automotive Engineering, Stopping Distance Calculations Help?

1. Oct 11, 2011

### eflickinger

1. The problem statement, all variables and given/known data

A vehicle weighs 3000 lbf and has an initial velocity of 68 mph. If the braking system is designed to stop the vehicle at constant power, calculate the stopping distance in feet, if the stopping power is 100 hp.

m = 3000 lbf / 32.2 ft/s2 = 93.17 (lbf-s2)/ft
v = 68 mph = 68 miles/hr x 5280 ft/mile x 1hr/3600s = 99.73 ft/s
P = 100 hp = 100 hp x 550 (ft-lbf)/s / 1 hp = 55,000 (ft-lbf)/s

2. Relevant equations

Fb = Braking Force
P = Energy / Time
Energy = 0.5*m*v2
Time (to stop) = t = (mv)/Fb
Distance (to stop) = x = 0.5*(m/Fb)*v2

3. The attempt at a solution

We need to calculate x, the stopping distance.
m and v are defined, however Fb is not.
We can find the time (to stop) based on the power = energy / time equation since we have both energy and power.
Therefore, Fb = (mv)/t

Plugging back into distance (x),

0.5*(m/((mv)/t))*v2

This makes x, the stopping distance = 0.5*t*v

t = Power / Energy
Energy = 0.5*m*v2
Energy = 0.5(93.17)(99.73)2 = 463,337.81 ft-lbf
Power = 55,000 (ft-lbf)/s <-- As stated previously
Therefore, t = 463,337.81 / 55,000 = 8.42 seconds

Now plugging back into the stopping distance formula we derived:

0.5(8.42s)(99.73ft/s) = 420.08 ft

But somehow this isn't the correct answer according to my online homework (which also doesn't supply the correct answer)?

Thank You,

Evan

2. Oct 12, 2011

### 256bits

I am not sure if I follow this.
The question states that the stopping power is a constant 100hp.
As you know Power = Force x Velocity,
so as the velocity decreases when the breaks are applied, the breaking force will have to increase to keep the breaking power constant. Have you taken that aspect of the problem into consideration?

3. Oct 24, 2011

### eflickinger

Thank you, it helped me think about it differently.

I used P = F x V => P = maV and integrated the acceleration with respect to distance and velocity to get the correct answer.