You are Here: Home >< Maths

# edexcel M1 - kinematics Watch

1. This is the question

A car is travelling along a straight horizontal road with constant acceleration. The car passes over 3 consecutive point A, B and C. AB = 100m, BC= 300m. At B the speed of the car is 14m/s and at C it is 20m/s. Find the acceleration and the tome taken for the car to travel form A to C.

-----------------------------------------------------------------------------------------------
I got 0.34 for acc which is correct. Then I used s = vt -0.5at^2 and got two answers. What would they actually mean in practice? Why would it pass C twice?

The book solution first found the value of u at A and the used that in 20 = u + 0.34t thus getting only one value for t which was the lower of the two solutions I got.

Thanks
2. You will frequently get 2 answers when you use but normally it's just a case of discrediting one of the answers (like if you get t<0).
3. (Original post by B_9710)
You will frequently get 2 answers when you use but normally it's just a case of discrediting one of the answers (like if you get t<0).
Thanks - I get that but on what grounds could I discredit the other value as it wasn't -ve. I suppose the smaller value MUST be right, logically, but is that all I could do?
4. (Original post by maggiehodgson)
This is the question

A car is travelling along a straight horizontal road with constant acceleration. The car passes over 3 consecutive point A, B and C. AB = 100m, BC= 300m. At B the speed of the car is 14m/s and at C it is 20m/s. Find the acceleration and the tome taken for the car to travel form A to C.

-----------------------------------------------------------------------------------------------
I got 0.34 for acc which is correct. Then I used s = vt -0.5at^2 and got two answers. What would they actually mean in practice? Why would it pass C twice?

The book solution first found the value of u at A and the used that in 20 = u + 0.34t thus getting only one value for t which was the lower of the two solutions I got.

Thanks

The equation s = vt -0.5at^2 is a quadratic in t, either "u" shaped or "n" shaped, and for a given displacment (s) you often have two values for t.

If the acceleration had remained constant throughout, then at some stage in the past (before t=0), the velocity would have been negative and the car would have been travelling backwards, first through C, then B, then A, eventually coming to a stop and then moving forward, and it's this latter part that the question is asking about.
5. (Original post by maggiehodgson)
Thanks - I get that but on what grounds could I discredit the other value as it wasn't -ve. I suppose the smaller value MUST be right, logically, but is that all I could do?
In this case the higher value relates to the time when the car first passed A, going backwards! Note there is no "u" in your equation. For the higher time u would have been negative, which is contrary to the question.

The difference between the two times is the time from when the car was going backwards through A (prior to t=0), until it came to a stop, and started going forwards again and got to A (at t=0).
6. (Original post by ghostwalker)
In this case the higher value relates to the time when the car first passed A, going backwards! Note there is no "u" in your equation. For the higher time u would have been negative, which is contrary to the question.

The difference between the two times is the time from when the car was going backwards through A (prior to t=0), until it came to a stop, and started going forwards again and got to A (at t=0).

Wow! Thanks
7. (Original post by maggiehodgson)
Wow! Thanks
Just to clarify, I was taking a few liberties with the definition of t there.

For the actual question t=0 refers to when the car was moving forwards at A, giving you the lower time.

The higher answer refers to t=0 when the car was moving backwards at A, hence the greater time.
8. (Original post by ghostwalker)
Just to clarify, I was taking a few liberties with the definition of t there.

For the actual question t=0 refers to when the car was moving forwards at A, giving you the lower time.

The higher answer refers to t=0 when the car was moving backwards at A, hence the greater time.

As usual I'm struggling to get this.

When it was moving backwards at A, where was it coming from? Was it moving in the direction B-A and if so where did it stop and turn round? And why did it stop because the acceleration was +ve.

I don't suppose you could give me a drawing could you?
9. (Original post by maggiehodgson)
As usual I'm struggling to get this.

When it was moving backwards at A, where was it coming from? Was it moving in the direction B-A and if so where did it stop and turn round? And why did it stop because the acceleration was +ve.

I don't suppose you could give me a drawing could you?
Here's a time displacement graph, set up for positive velocity at A at t=0.

Displacment is on the vertical axis, and time on the horizontal.

Edit: Since acceleration is constant, displacment against time is a parabola (or part of one) - as is always the case with suvat. It will be inverted if the acceleration is negative.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: May 26, 2016
Today on TSR

### Does race play a role in Oxbridge admissions?

Or does it play no part?

### The worst opinion about food you'll ever hear

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.