# M1 graph question.Watch

#1
I am currently trying to do this question, however we never got taught on how to do it, I have done part A, but I am stuck on part B and C.

The question is as follows:

A sprinter runs a race of 200m. Her total time for running is 25s. Figure 1 is a sketch of the speed-time graph for the motion of the sprinter. She starts from rest and accelerates uniformly to a speed of 9m/s in 4s. The speed 9m/s is maintained for 16s and she then decelerates uniformly to a speed of um/s at the end of the race. Calculate:

(a) the distance covered by the sprinter in the first 20s of the race,
(b) the value of U,
(c)( the deceleration of the sprinter in the last 5s of the race.

For part (a) is got: 162m

I cannot do part (b) and would you do part (c) with the equation of motion or something else?
0
9 years ago
#2
...
You've not said how you did part a).

Are you familiar with the fact that the area under a velocity time graph is the distance covered in that time? If you weren't, well, you are now.

Part b)
Considering the final trapezium. What's its area in terms of "u"? And what is the distance covered during this time (you can work that out from your answer to a)?
#3
part a) I did the follwing:

area under graph:

A = 1/2bh (triangle bit) + bh (rectangle bit of graph).

A = 18+144 = 162m

or is that wrong?

using the area of trapezium also gives me the answer 162m.
0
9 years ago
#4
part a) I did the follwing:

area under graph:

A = 1/2bh (triangle bit) + bh (rectangle bit of graph).

A = 18+144 = 162m

or is that wrong?
That's correct.

Just apply the same principle to the last bit, except you'll have an unknown quantity in the area, "u". And you can then solve the equation for u.
#5
=/ finding U is proving difficult for me, I keep getting 9m/s, which cannot be right,
0
9 years ago
#6
=/ finding U is proving difficult for me, I keep getting 9m/s, which cannot be right,
Post some working, so we can see what's happening. Area of a trapezium would be easiest, but if you are not familiar with that, then use the sum of the rectangle and triangle. Just on the last section of t=20 to 25.
#7
yeah for the trapezium rule I get:

1/2(a+b)h
1/2(9+u)5 = 5/2(9+u)

but what do you equal it to?
0
9 years ago
#8
yeah for the trapezium rule I get:

1/2(a+b)h
1/2(9+u)5 = 5/2(9+u)

but what do you equal it to?
It's a 200m race and the sprinter has covered 162 m prior to this final interval.

So....
#9
ahh right, I didn't see that bit XD

0
9 years ago
#10
That's what I make it.

And the final part will just be change in velocity divided by time.
#11
so the deceleration is -0.112 ?
0
9 years ago
#12
so the deceleration is -0.112 ?

What's the change in velocity in that final section?
How long does it take?
#13
sorry xD is it (9-6.2)/5 = 0.56?
0
9 years ago
#14
sorry xD is it (9-6.2)/5 = 0.56?
Bingo. That's the deceleration.
9 years ago
#15
ahh right, I didn't see that bit XD

Sorry to bump up this thread but can i ask how you got from; 1/2(a+b)h
1/2(9+u)5 = 5/2(9+u)

to 6.2?

is it 5/2(9+u) = 38?
or have i gone wrong?

i seem to have gotten myself a little lost
0
9 years ago
#16
(Original post by SoothsayerXYZ)
is it 5/2(9+u) = 38?
Yes. 200-162=38
9 years ago
#17
(Original post by ghostwalker)
Yes. 200-162=38
so how did you get to 6.2 from there?
(from here);
5/2(9+u) = 38

i make that

9 + u = 95

9 - 95 = u

u = -86

which is obviously wrong for many reasons.
0
9 years ago
#18
(Original post by SoothsayerXYZ)
so how did you get to 6.2 from there?
(from here);
5/2(9+u) = 38

i make that

9 + u = 95

9 - 95 = u

u = -86

which is obviously wrong for many reasons.
9+u = 38*2/5=15.2
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