Calculating Intercept Time and Distance in a Police Chase

[m00c0w]

Hi I'm extremely sorry if this is the wrong forum but it seemed the sensible choice. Here goes:

"A stolen car, traveling at a constant speed of 40m/s, passes a police car parked in a lay-by. The police car sets off three seconds later, accelerating uniformly at m/s². How long does the police car take to intercept the stolen vehicle and how far from the lay-by does this happen?"

I got answers of 3.94s and 320m. Could anyone confirm if I am right or not, and if not I'd be grateful if someone could suggest a method of doing it correctly.

Many thanks!

 

[brewnog]

Well you didn't say the value of the police car's acceleration...

 

[m00c0w]

Sorry I mistyped.. the police car accelerates uniformly at 8m/s².

 

[Fermat]

Your answers are wrong, I'm afraid.

How did you do this ?
Could you show some working ?

 

[BobG]

Hi I'm extremely sorry if this is the wrong forum but it seemed the sensible choice. Here goes:
"A stolen car, traveling at a constant speed of 40m/s, passes a police car parked in a lay-by. The police car sets off three seconds later, accelerating uniformly at m/s². How long does the police car take to intercept the stolen vehicle and how far from the lay-by does this happen?"
I got answers of 3.94s and 320m. Could anyone confirm if I am right or not, and if not I'd be grateful if someone could suggest a method of doing it correctly.
Many thanks!

Your answer is wrong and it's hard to imagine how you came with them (you didn't seem to make the most likely mistake). How did you come up with those numbers? You should be able to take the scenario given and form a pretty standard kinematic equation (s_f = s_i + vt + 1/2 at^2).

Given a constant velocity of the stolen car, you can calculate the initial position of the stolen car at the time the police car started to move. At some point in time, both cars will occupy the same space. Rearranging the equation yields the kinematic equation in the first paragraph.

 

[m00c0w]

Firstly I worked out how long it takes the police car to match the speed of the stolen car. I got an answer of 5 seconds using v = u + at. I then worked out that it traveled 100m in these 5 seconds using s = 1/2(u+v)t. I calculated how far the stolen car traveled in 8 seconds (3 seconds the police car waited and the 5 seconds it took to match the stolen car's speed) by 8 x 40m/sec = 320m. I worked out the distance of the 2 cars to be 220m and used s = ut + 1/2at² to find out how long it took the police car to catch up with the stolen car. I've realized I forgot to add the initial 8 seconds to 3.94 but I'm assuming that 11.94 is still the wrong answer?

 

[finchie_88]

The answers I got were:
(Written in white text, so that if you don't want to know, don't highlight the following text):
Distance = 400 metres, and time = 10 seconds
Tell me if anyone else thinks these are right (or wrong), because it would be helpful to me if I knew.

 

[Fermat]

The answers I got were:
(Written in white text, so that if you don't want to know, don't highlight the following text):
Distance = 400 metres, and time = 10 seconds
Tell me if anyone else thinks these are right (or wrong), because it would be helpful to me if I knew.

Still wrong, I'm afraid
When you want to check on your ansers, could you also post your working ?
That way, we can point out just where your error(s) are when replying to your post. Thanks.

 

[Fermat]

Firstly I worked out how long it takes the police car to match the speed of the stolen car. I got an answer of 5 seconds using v = u + at. I then worked out that it traveled 100m in these 5 seconds using s = 1/2(u+v)t. I calculated how far the stolen car traveled in 8 seconds (3 seconds the police car waited and the 5 seconds it took to match the stolen car's speed) by 8 x 40m/sec = 320m. I worked out the distance of the 2 cars to be 220m and used s = ut + 1/2at² to find out how long it took the police car to catch up with the stolen car. I've realized I forgot to add the initial 8 seconds to 3.94 but I'm assuming that 11.94 is still the wrong answer?

OK, thanks for posting your working, I can see what you're doing and where you're going astray.
When you got the distance between the stolen car and the police car as 220m, you then used s = ut + (1/2)at² to find out how long it would take the police car to catch up. That was where the error is. By the time the police car travels that 220m, the stolen car has moved on again! (by about 3.94*40 = 157m) You've gotten yourself into a hare and tortoise scenario. Every time the police car/hare catches up with the stolen car/tortoise, it has moved on!

To solve this problem ...

Both cars will have covered the same distance (when the police car finally catches up) but they will take different times (measured from when the stolen car passes the police car)

Let T secs be the time the stolen car travels before being caught up (its journey time)
Then the journey time of the police car is 3 seconds less, i.e. (T-3) secs.
How far has the stolen car traveled in T secs ?
How far has the police car traveled in (T-3) secs ?
Equate the two distances.

 

[m00c0w]

You should be able to take the scenario given and form a pretty standard kinematic equation (s_f = s_i + vt + 1/2 at^2)

Could someone please explain what s_f and s_i are?

 

[Fermat]

S_final and S_initial

 

[BobG]

S_final and S_initial

... with s being position.