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# Urgent: Sg physics 2002 credit q10 Watch

1. I do not know how to do this question , this is important as its the only part of the course that im not sure of , anyone who has the paper plaese help
2. What's the question?
3. (Original post by Delaney)
What's the question?
its a question from unit 5 transport, and its got a graph so i cant really write it down as i will need to refer to the graph
4. Maybe you can tell us the question and someone would be able to tell you how to iterpret the graph in relation to the question? A desc of the graph could be good too.
5. (Original post by Ecosse_14)
Maybe you can tell us the question and someone would be able to tell you how to iterpret the graph in relation to the question? A desc of the graph could be good too.
i will scan the question 2morrow as i cant now so you can see it.
6. (Original post by animelover123)
i will scan the question 2morrow as i cant now so you can see it.
Message me when you have scanned it, i will help you to understand it.
Message me when you have scanned it, i will help you to understand it.
The file is too big to put on the internet
8. (Original post by animelover123)
The file is too big to put on the internet

Take a screenshot of it, and upload the picture
Take a screenshot of it, and upload the picture
oK IT MIGHT NOT BE CLEAR
[ATTACH][/ATTACH]
Attached Images

10. (Original post by animelover123)
oK IT MIGHT NOT BE CLEAR
[ATTACH][/ATTACH]

Sorry , its the first time ive attched afile in this fourm...shame.... made amess of it , its basicallly the 2 hyperlinks with name untitled
11. 10. a) Is the point the greyhound line crosses the x-axis as 0 is when the hare crosses the line.

b) Do you use the formula a = (v-u) / t in SG physics? If so, try and use that formula to find the acceleration.
12. (Original post by animelover123)
Sorry , its the first time ive attched afile in this fourm...shame.... made amess of it , its basicallly the 2 hyperlinks with name untitled

Part a)

Read the question carefully so you understand what is happening.

It tells you that the greyhounds are released AFTER the hare crosses the start line. So the hare starts FIRST.

It also tells you that the graph represents the motion when they CROSS THE START LINE.

Therefore the hair starts at 0 seconds, as it is the graph of when the hare starts.

IF you look closely, you can see that 0.5s later, the greyhounds start. Because each box represents 0.5seconds.

Part b)

We know that acceleration = v - u / t

From the graph, we notice that the greyhound starts with a sloped line. This sloped line represents an increase in speed, so it represents the acceleration.

V= Final speed, which if you look at the end of the sloped line, is 13m/s. ( as it becomes a straight line, this means the speed is now constant, so this is the final speed for our acceleration )

u=Initial speed, which we can see from the graph is 0

t= time, which is 6.5s, as if u look, from the start to the end of the sloped line, it is 6.5s

So therefore.

a= 13-0/6.5

acceleration = 2m/s/s

Part c)

The question tells us that the hare crosses the finish line 20s after starting.

to work out distance, we us d=vt.

distance = ?

v = speed , which we know is constantly 12m/s

t= time, which we know is 20s

so use d=vt

d = 12 x 20
d=240m.

Hope that helps
13. Part c)ii)

You need to know how far the greyhound has travelled until 20s.

To do this, work out the area under the greyhounds graph.

So first work out the triangle, which is 13 x 6.5 / 2 = 42.25m

Then work out the rest of the area.

The difference from the start of the rectangle to the 20 is 13 seconds, therefore;

13 x 13 = 169.

169 + 42.25 = 211.25

Hare - hound

= 240 - 211.25
= 28.75 / 29m

Hopefully this helps..
Please someone correct me if im wrong, i cant quite remember if this is right!

You need to know how far the greyhound has travelled.

To do this, work out the area under the greyhounds graph.

So first work out the triangle, which is 13 x 6.5 / 2 = 42.25m

The hare finishes 20s after it starts. The hound we shall messure of 19.5s. You must notice that the hound doesn't actually go anywhere for 0.5s. So after 20s it has ONLY MOVED for 19.5s. As it was stationary for the first 0.5s.

We have already worked out the first 6.5s, so we take that away from our total journey time of 19.5s, and that gives us 14 seconds for the rest of the graph.

Then work out the rest of the area.

14 x 3 = 182.

182 + 42.5 = 224.5

Hare - hound

= 240 - 224.5

= 15.5m

Basically, your working out the area under the graph , up until 20 seconds. So if your looking at the graph you will automatically notice that the time from where the graph starts ( 0.5 ) to where we are finishing ( 20 s ) the time is 19.5s

Hopefully this is right... :O
Thanks very much for answering much appreciated, the past papers don't tell you the working so its quite hard if you do not know what to do , but i have checked and your last answer is incorrect the answer states

(cii) 28.75m or 29 m

so i dont know if you can tell me how to get there
15. (Original post by animelover123)
Thanks very much for answering much appreciated, the past papers don't tell you the working so its quite hard if you do not know what to do , but i have checked and your last answer is incorrect the answer states

(cii) 28.75m or 29 m

so i dont know if you can tell me how to get there
Check it now.

Because the picture was quite shady, i read from 6 to 20 instead of 7 to 20 !

Look at it now
Check it now.

Because the picture was quite shady, i read from 6 to 20 instead of 7 to 20 !

Look at it now
IO see it now i was looking at the latter post but you put it into the your 1st answer post. But thanks anyway
17. Oops at not realising the attachments weren't the same thing
18. (Original post by Ecosse_14)
Oops at not realising the attachments weren't the same thing
hah.. that was kinda my fault... ****ty computer ...lol

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