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# Can anyone help me with this question? watch

1. I really can't do this question can anyone give me some tips? Here's the question: a car is travelling along a straight horizontal road. The car take 120s to travel between two sets of traffics lights with are 2145m apart. The car starts from rest at the first set of traffic lights and moves with constant acceleration for 30s until it's speed is 22ms. The car maintains this speed for t seconds. The car the moves with constant deceleration, coming to a rest at the second set of traffic lights. Find the value of t.
3. Has someone replied?I cannot see it but got a notification for it.
4. (Original post by Jiffylemmon)
I really can't do this question can anyone give me some tips? Here's the question: a car is travelling along a straight horizontal road. The car take 120s to travel between two sets of traffics lights with are 2145m apart. The car starts from rest at the first set of traffic lights and moves with constant acceleration for 30s until it's speed is 22ms. The car maintains this speed for t seconds. The car the moves with constant deceleration, coming to a rest at the second set of traffic lights. Find the value of t.
I think a speed-time graph would help for this question. Then use the fact that the area under the graph gives you the total distance which is 2145.

Have a go at that and post where you got up to if you're still stuck.
5. (Original post by notnek)
I think a speed-time graph would help for this question. Then use the fact that the area under the graph gives you the total distance which is 2145.

Have a go at that and post where you got up to if you're still stuck.
Erm that is the problem that we haven't be told how to do the time graph, and are teacher told us not to use one.
6. (Original post by Jiffylemmon)
Erm that is the problem that we haven't be told how to do the time graph, and are teacher told us not to use one.
OK then divide the distance into three regions and calculate the distance separately for each. For the first region you can use to find the distance.

You can work out the distance for the middle region in terms of t.

For the final region, since the total time is 120, the time for the final region must be (120 - t - 30). Use this to find the distance here in terms of t.

By the way, speed-time graphs are part of M1 and would make this question quite simple.
7. (Original post by notnek)
OK then divide the distance into three regions and calculate the distance separately for each. For the first region you can use to find the distance.

You can work out the distance for the middle region in terms of t.

For the final region, since the total time is 120, the time for the final region must be (120 - t - 30). Use this to find the distance here in terms of t.

By the way, speed-time graphs are part of M1 and would make this question quite simple.
Yeah, I have seen how to an time graph question exactly like this with number changed. Can I ask I have the answer 57s but I don't know how I got to it. Also for the first distance I have 330 m and the acceleration is 11 over 15.
8. (Original post by Jiffylemmon)
Yeah, I have seen how to looked at time graph question exactly like this with number changed. Can I ask I have the answer 57s but I don't know how I got to it.
57s is incorrect. Post your working.
9. (Original post by Jiffylemmon)
I really can't do this question can anyone give me some tips? Here's the question: a car is travelling along a straight horizontal road. The car take 120s to travel between two sets of traffics lights with are 2145m apart. The car starts from rest at the first set of traffic lights and moves with constant acceleration for 30s until it's speed is 22ms. The car maintains this speed for t seconds. The car the moves with constant deceleration, coming to a rest at the second set of traffic lights. Find the value of t.
This exact question is from a 2013 Edexcel M1 paper. Part a) of the question asks you to draw a speed-time graph.
10. (Original post by notnek)
57s is incorrect. Post your working.
Okay so I have (0+22/2) x 30 for the first distance of 330m and I don't understand how you're suppose to put the 2nd and 3rd distance in terms of t?
11. (Original post by Jiffylemmon)
Okay so I have (0+22/2) x 30 for the first distance of 330m and I don't understand how you're suppose to put the 2nd and 3rd distance in terms of t?
The car travels at a constant speed of 22m/s for t seconds. How far does it travel in terms of t?

Spoiler:
Show
s=ut+0.5at^2 or just speed = distance/time since there is no acceleration
12. (Original post by notnek)
This exact question is from a 2013 Edexcel M1 paper. Part a) of the question asks you to draw a speed-time graph.
Yes, I know however like I have said my teacher said we should miss out on it and try to do it without one since we are learning about time graphs next week. (On a side note; is this a question that I should really be getting in my first week of college, I haven't done much maths in the pass 2 months.)
13. (Original post by Jiffylemmon)
Yes, I know however like I have said my teacher said we should miss out on it and try to do it without one since we are learning about time graphs next week. (On a side note; is this a question that I should really be getting in my first week of college, I haven't done much maths in the pass 2 months.)
OK. It's useful to know that the question was designed for using speed-time graphs. So the method you're trying to use now is not the one that you should use in the future.
14. (Original post by notnek)
The car travels at a constant speed of 22m/s for t seconds. How far does it travel in terms of t?
Spoiler:
Show
s=ut+0.5at^2 or just speed = distance/time since there is no acceleration
I'm really sorry but I still really don't know. I feel like a right idoit right now.
15. (Original post by Jiffylemmon)
I really sorry but I still really don't know. I feel like a right idoit right now.
It's just a case of plugging numbers (and letters) into a formula. You can just use distance = speed x time since there is no acceleration.

Speed = 22
Time = t

Post a guess at what you think the distance is if you're still not sure.
16. (Original post by notnek)
It's just a case of plugging numbers (and letters) into a formula. You can just use distance = speed x time since there is no acceleration.

Speed = 22
Time = t

Post a guess at what you think the distance is if you're still not sure.
So is it 2nd distance-3rd distance -1st distance (330)/s (22)=t?
17. (Original post by Jiffylemmon)
So is it 2nd distance-3rd distance -1st distance (330)/s (22)=t?
I don't know what any of that means.

Are you trying to find the distance in the middle region? Or something else?
18. (Original post by notnek)
I don't know what any of that means.

Are you trying to find the distance in the middle region? Or something else?
Yes I'm trying to find the middle region, however I don't know the last/ deceleration region. If I knew the last region I would be able to do it. Since I already have the first region figured out and already have the constant speed stated in the question.
19. (Original post by Jiffylemmon)
Yes I'm trying to find the middle region, however I don't know the last region. If I knew the last region I would be able to do it. Since I already have the first region figured out and already have the constant speed stated in the question.
Ignore the final region for now and concentrate on the middle region. What is the distance travelled for the middle region in terms of t?
20. (Original post by notnek)
Ignore the final region for now and concentrate on the middle region. What is the distance travelled for the middle region in terms of t?
22xs=t? The constant speed it the only information that I know that relates to the question?

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