# Thread: Calculating the braking distance with equations

1. ## 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

2. 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.

3. ## 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.