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# M1 question. .HEEEELP! watch

1. Hi. .i missed a couple lessons, and am now so stuck on my homework. It's probably really easy but. .

A boy is walking at a constant speed of 1.8m/s along a straight road. He passes a telephone booth where his sister is making a call. His sister takes 30s to complete the call and the sets off in pursuit of the boy. She accelerates uniformly from rest at 3m/s*2 until she's running at a speed of 9m/s. She maintains this constant speed until she reaches her brother.

Q. Calculate the time taken by the girl to reach her brother?

2. ok, so calculate the distance the boy goes (you know his speed) during his sister's 30s phone call. That's the distance she has to cover. She is accellerating at 3m/s^2 up to 9m/s then stays at constant speed. Just use one of the suvat equations (involving what you know about the girl, ie a, u and v) to find t

Can you take it from there?
3. Well because it asks you for the time taken when she reaches her brother, you know their displacement will be the same at that point. So you derive expressions for the distance travelled by both of them. If you assume the total time taken is (30+t) to account for the time for sister to complete the call then:

Distance travelled by boy = 1.8(30+t)

Distance travelled by girl = (0.5*3*3*3)+(t-3)9

You then make them equal to each other and solve for "t".

(Note: in the second equation I used "s = ut + 0.5at^2" to find the distance travelled while she is accelerating and used t=3 because the time to reach 9ms^-1 while accelerating at 3ms^-2 while be 3s.
For the (t-3)9 bit, its her speed "9" times by the time (t-3). (t-3) was used because of the time she was accelerating for.)

Sorry if it sounds confusing
4. I would do it by drawing a velocity-time graph showing both journeys on the same diagram.

The boy's journey would form a rectangle.

The girls journey would form a trapezium.

The area of both shapes represents the distance travelled and would be equal.
5. (Original post by Plato's Trousers)
ok, so calculate the distance the boy goes (you know his speed) during his sister's 30s phone call. That's the distance she has to cover. She is accellerating at 3m/s^2 up to 9m/s then stays at constant speed. Just use one of the suvat equations (involving what you know about the girl, ie a, u and v) to find t

Can you take it from there?
but doesn't the boy carry on moving, so she has a distance of further than 54m to run?
6. (Original post by Anon1993)
but doesn't the boy carry on moving, so she has a distance of further than 54m to run?
Yes. The distance covered by the boy is 1.8t where t is the time you are trying to find.
7. It does not say that the boy stops after finishing the phone call. So she will have to cover the distance travelled in the phone call as well as the distance the boy has travelled after the phone call. It may make it easier to draw a couple of speed-time graphs to make it easier to visualise. The area under the graph is the distance, so the total area under both graphs must be equal.
8. (Original post by Mr M)
I would do it by drawing a velocity-time graph showing both journeys on the same diagram.

The boy's journey would form a rectangle.

The girls journey would form a trapezium.

The area of both shapes represents the distance travelled and would be equal.
Ahh gotchya. That makes it embarrassingly easy! I got an answer of 39.375 s?
9. (Original post by soutioirsim)
Well because it asks you for the time taken when she reaches her brother, you know their displacement will be the same at that point. So you derive expressions for the distance travelled by both of them. If you assume the total time taken is (30+t) to account for the time for sister to complete the call then:

Distance travelled by boy = 1.8(30+t)

Distance travelled by girl = (0.5*3*3*3)+(t-3)9

You then make them equal to each other and solve for "t".

(Note: in the second equation I used "s = ut + 0.5at^2" to find the distance travelled while she is accelerating and used t=3 because the time to reach 9ms^-1 while accelerating at 3ms^-2 while be 3s.
For the (t-3)9 bit, its her speed "9" times by the time (t-3). (t-3) was used because of the time she was accelerating for.)

Sorry if it sounds confusing
Haha confusing, but i think i got you. .

You know the 0.5*3*3*3 bit, is that the same as the area of the acceleration 'triangle' if its drawn as a graph?
10. (Original post by Anon1993)
Ahh gotchya. That makes it embarrassingly easy! I got an answer of 39.375 s?
Yup.
11. (Original post by Anon1993)

You know the 0.5*3*3*3 bit, is that the same as the area of the acceleration 'triangle' if its drawn as a graph?
Exactly! The graph should look like this...
Attached Images

12. (Original post by soutioirsim)
Exactly! The graph should look like this...
Woo miracle of miracles, i actually got that graph!
Thanks. . .reps tmw
13. shouldn't the time spent accelerating be 9s not 3s?
14. (Original post by fbc)
shouldn't the time spent accelerating be 9s not 3s?
No.

Use v = u + at with v = 9 m/s, u = 0 m/s and a = 3 m/s^2 to check.

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