Finding the distance from deceleration

[Marstiphal]

Homework Statement

A lorry is driving at 96 km h^−1. The driver decelerates to a speed of 48 km h^−1. Given that the deceleration is 2.68 m s^−2 find the distance over which the brakes are applied.

I know the initial speed of the lorry, v = 96 km h^-1 = 26.67 m s^-1
I know the final speed of the lorry, u = 48 km h^-1 = 13.33 m s^-1
I know the deceleration rate of the lorry, a = 2.68 m s^-2

so I'm looking for d (Distance)

Homework Equations

t=(v-u)/a to find the time t the breaks were applied for

d = vt + 1/2 x at^2 ?? to find the distance traveled while decelerating?

The Attempt at a Solution

Substitute the values for v,u and a to find speed s
t=(26.67 m s^-1 - 13.33 m s^-1)/2.68 m s^-2

= 4.98 s

I'm not sure what the second equation is for working out distance the lorry has traveled based on a decelerating body and the info I now have.

so I think d= (26.67 m s^-1 x 4.98 s) + 1/2 x 2.68 m s^-2 x 4.98 s^2?

Also, if this is correct and based on the information that I have, is there any other equation I could use to also find distance? I ask as the question's hint tells me to consider the energies involved and the section just read is also related to energy and force...

Many thanks

 

[kuruman]

You can use energy considerations or you can use the kinematic equation

2 a (x_final - x_initial) = v_final² - x_initial²

which is essentially the same thing as the work-energy theorem. Don't forget that the acceleration is a negative number here.

 

[Marstiphal]

Hi Kuruman and thankyou for your answer. I'm not sure I totally understand the equation though. Is this how I would find distance based on the information I have? I'm not totally sure how I would use the work-energy theorem to solve this perticular problem either.

 

[kuruman]

Hi Kuruman and thankyou for your answer. I'm not sure I totally understand the equation though. Is this how I would find distance based on the information I have?

Take the first relevant equation that you quoted,
t = (v_final - v_initial)/a
and substitute for t in
d = v_initialt + (1/2)at².
If you do the algebra correctly, you should get the equation that I quoted. You should use this equation when you don't know (and don't care to know) what the time is.

I'm not totally sure how I would use the work-energy theorem to solve this perticular problem either.

The work-energy theorem says
W_Net = ΔK = (1/2)m(v_final² - v_initial²)
Now the net work is the work done by all the force and can be though of as the work done by the net force. If the displacement is d,
W_Net = F_Netd
and since F_Net = ma, W_Net = mad. So,
mad = (1/2)m(v_final² - v_initial²)
from which you get
2ad = (v_final² - v_initial²).