# Motion in 1 and 2 dimensions

#### chaos17

1. Homework Statement

In a road test,a car was uniformly accelerated from rest over a distance of 400m in 18.5s. The driver then applied the brakes, stopping the car in 4.9 s with constant deceleration.

a) calculate the acceleration of the car for the first 400m.

b)Calculate the average speed of the car for the journey covering both the acceleration and braking distances.

2. Homework Equations

v=d/t

avg speed=distance / time

3. The Attempt at a Solution

a)v=d/t

v=400/18.5
=21.62ms^-1

b) I have no idea, I don't know what to do. Could someone give me a hint on what values to use or what formula to use.

Is my part a right?

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#### chaos17

what i mean by part a being right is that do i then use v to work out acceleration? Or is this a wrong way?

#### The Liberator

From what I can see, part (a) is right. The first thing that would be good to do is, to split the acceleration and the deceleration sections up. As you know the velocity of the first part, you need to find the velocity of the second part. How would you go about doing that, as you have the time and the initial velocity of the car, but you do not have the deceleration, final velocity or the distance the deceleration acts over. You do not need to know all everything, just enough to find out the average velocity over the whole[b/] time.

In reply to second post by Chaos: I believe you are on the right track. What equation do you use to then work out acceleration, when you have the velocity, time and distance?

Last edited:

#### chaos17

I would them use avg acceleration= change in velocity / time.

Now that I think of it, wouldn't part a be:

u=0, t=18.5, x=400

x=ut+0.5at^2

so 400=0.5 * a * (18.5^2)
a=400/171.125
a=2.34m^-2

I think that is right.

But them what do I do for part b.

#### chaos17

can anyone help for part b?

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