# Calculating the braking distance with equations

• Mar 9th 2010, 11:45 PM
jamesgill
Calculating the braking distance with equations
Hi,
Can anybody help me with the following as I am totally stuck! A truck is moving at 88 km h-1 and breaks to 52km h -1. What is the distance over which the breaks are used if the deceleration was 2.82 m s -2.

I have to use one or more of the following equations - and correct SI units. I have been told to think about the energies involved. Any help will be gratefully received.

Ek = 1/2mv^2

W = Fd

P = E/t
• Mar 10th 2010, 01:49 AM
HallsofIvy
Quote:

Originally Posted by jamesgill
Hi,
Can anybody help me with the following as I am totally stuck! A truck is moving at 88 km h-1 and breaks to 52km h -1. What is the distance over which the breaks are used if the deceleration was 2.82 m s -2.

I have to use one or more of the following equations - and correct SI units. I have been told to think about the energies involved. Any help will be gratefully received.

Ek = 1/2mv^2

W = Fd

P = E/t

I would have thought of using a motion equation but the Energy equation you have here works nicely. At 88 km/hr, the truck has kinetic energy of \$\displaystyle (1/2)m(88)^2\$ Joules. We don't know the trucks mass but wait on that.

At 52 km/h, its kinetic energy is \$\displaystyle (1/2)m(52)^2\$ Joules. The work done by the brakes is the difference between those two: \$\displaystyle (1/2)m(52)^2- (1/2)m(88)^2= (1/2)m(52^2- 88^2)\$ Joules.
(Notice the order of those speeds. The truck slows down so the work done is negative.)

The truck decelerates at \$\displaystyle 2.82 m/s^2\$ so its acceleration is \$\displaystyle -2.82 m/s^2\$. Force equals mass times acceleration (that formula isn't given here but I am sure you know it): \$\displaystyle F= ma= -m(2.82 m/s^2)\$.

Now use W= Fd: \$\displaystyle (1/2)m(52^2- 88^2)= -m(2.82 m/s^2)d\$.

The unknown mass, m, cancels and you can solve for d.
• Mar 10th 2010, 02:51 AM
jamesgill
Thank you
Thank you so much for not just the answer but also the clear explanation. I should be able to tackle similar equations myself now.