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# When will 2 cars meet given the velocity of one and the acceleration of the other? watch

1. Got a question here...can't work out whether the answer I've got is wrong, or if the answer on the sheet is wrong!

A stolen car is travelling at a constant velocity of 40m/s which it overtake a stationary police car. Two seconds later the police car sets off in pursuit, accelerating uniformly at
How far does the police car travel, to the nearest metre, before catching up with the stolen car?

Here's my working:
Let t be the time before they meet.
So to get the distance for the stolen car, it would be 40t + 80 (as it had a 2 second head start)
To go from acceleration to distance, you multiply by time squared, so the distance for the police car is

You want to find the time when the distance is equal from the starting point, so you equate the equations

So therefore,
Solve for t and you get
If you plug that back into either and I get 683.976 which to the nearest metre is 684m.

The answer given is 1221m but I can't see how to get that at all...
If someone could give me a poke in the right direction that would be appreciated
2. (Original post by Lemur14)
To go from acceleration to distance, you multiply by time squared, so the distance for the police car is
Not quite, note that so you forgot the half.

The rest is correct procedure.
3. (Original post by RDKGames)
Not quite, note that so you forgot the half.

The rest is correct procedure.
Ohh were we meant to use suvat equations? We haven't got to them yet (although I briefly did them in add maths last year) Trust my school to put a suvat question on the velocity time graph homework
Thanks will take another look with that equation now
4. (Original post by Lemur14)
Ohh were we meant to use suvat equations? We haven't got to them yet (although I briefly did them in add maths last year) Trust my school to put a suvat question on the velocity time graph homework
Thanks will take another look with that equation now
In the exam you would tackle this with SUVAT, definitely.

Having said that, I'm not sure what you've covered to give an alternative way of doing it.
5. (Original post by RDKGames)
In the exam you would tackle this with SUVAT, definitely.

Having said that, I'm not sure what you've covered to give an alternative way of doing it.
Ok, thanks
We've literally had one lesson on displacement time graphs and one on velocity time graphs, so haven't really got beyond working out what the gradient means on each of them and what the area means, so that's why I took the approach of acceleration to distance is multiplied by time squared! It's the last question on the sheet though, so I suspect the school were just trying to stretch us!
6. (Original post by RDKGames)
In the exam you would tackle this with SUVAT, definitely.

Having said that, I'm not sure what you've covered to give an alternative way of doing it.
You could do it with vel-time graphs if you were told you had to.
7. (Original post by Lemur14)
Ok, thanks
We've literally had one lesson on displacement time graphs and one on velocity time graphs, so haven't really got beyond working out what the gradient means on each of them and what the area means, so that's why I took the approach of acceleration to distance is multiplied by time squared! It's the last question on the sheet though, so I suspect the school were just trying to stretch us!
Have you tried drawing a velocity time graph with both cars on the same axes and considered areas? It sounds like that's the expected approach.
8. (Original post by Lemur14)
Got a question here...can't work out whether the answer I've got is wrong, or if the answer on the sheet is wrong!

A stolen car is travelling at a constant velocity of 40m/s which it overtake a stationary police car. Two seconds later the police car sets off in pursuit, accelerating uniformly at
How far does the police car travel, to the nearest metre, before catching up with the stolen car?

Here's my working:
Let t be the time before they meet.
So to get the distance for the stolen car, it would be 40t + 80 (as it had a 2 second head start)
To go from acceleration to distance, you multiply by time squared, so the distance for the police car is

You want to find the time when the distance is equal from the starting point, so you equate the equations

So therefore,
Solve for t and you get
If you plug that back into either and I get 683.976 which to the nearest metre is 684m.

The answer given is 1221m but I can't see how to get that at all...
If someone could give me a poke in the right direction that would be appreciated
Once you've drawn the graph you'll see where the 1/2 comes from that RDKGames mentioned. And maybe your teacher will use this as a nice intro to SUVAT
9. (Original post by Muttley79)
You could do it with vel-time graphs if you were told you had to.
I tried and I got confused, but thanks!
(Original post by Notnek)
Have you tried drawing a velocity time graph with both cars on the same axes and considered areas? It sounds like that's the expected approach.
Yeah I drew a graph but since I didn't know the greatest velocity of the police car I wasn't really sure how to go from the diagram, hence I tried this!
(Original post by Notnek)
Once you've drawn the graph you'll see where the 1/2 comes from that RDKGames mentioned. And maybe your teacher will use this as a nice intro to SUVAT
I can hope...this teacher doesn't seem to do nice intros from one thing to another...we just do a lot of jumping around of topics...apparently she's saving one last lesson of pure for us to do at some point...we had an empty lesson before half term it could have filled perfectly
10. (Original post by Lemur14)
Yeah I drew a graph but since I didn't know the greatest velocity of the police car I wasn't really sure how to go from the diagram, hence I tried this!
You don't need the greatest velocity. You'll have a horizontal line for the normal car and a diagonal line starting from t = 2 for the police car. There will be a point where they intersect and this will be where the police car has caught up in speed but not actually caught up with the car.

After this there will be a point where both the car and police car have travelled the same distance, mark this as a random point after the intersection. This is where the police car has caught up.

For this point you'll have an unknown x-axis length which you can call t. Then make equations for the total distance of the car and total distance of the police car and set them equal to each other.

11. (Original post by Notnek)
You don't need the greatest velocity. You'll have a horizontal line for the normal car and a diagonal line starting from t = 2 for the police car. There will be a point where they intersect and this will be where the police car has caught up in speed but not actually caught up with the car.

After this there will be a point where both the car and police car have travelled the same distance, mark this as a random point after the intersection. This is where the police car has caught up.

For this point you'll have an unknown x-axis length which you can call t. Then make equations for the total distance of the car and total distance of the police car and set them equal to each other.

Aha! Finally got it, thank you
Why do I always miss such stupid things

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