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# P3 Differential Equation watch

1. A girl live 500 metres from school. She sets out walking at 2ms , but when she has walked a distance of x metres her speed has dropped to ( 2 - 1/400 x ) ms. How long does she take to get to school?
2. have u got an answer, before i scare you with a mahoosive wrong solution?

thanks for helping!
4. No i got something like 9.5 minutes.. grr.

To be honest with you though, i'm using M2 methods.. so errrr, i'll keep trying and see what i come up with.

I'm basing it on the fact that INT (velocity) = displacement
5. (Original post by Sang)
A girl live 500 metres from school. She sets out walking at 2ms , but when she has walked a distance of x metres her speed has dropped to ( 2 - 1/400 x ) ms. How long does she take to get to school?
when distance = 0 then speed = 2 ms
when distance = x then speed = 2 - 1/ 400x

time = distance / speed

x / 2 - 1/400 x = 400x^2 - 2 = x^2 = 0.005

so x = square root of 0.005

(actually thats probably wrong, im not sure.)
6. i got 6.53878 minutes. So i think you're right
7. (Original post by SUKBarracuda)
i got 6.53878 minutes. So i think you're right
8. dx/dt

dt/dx

=> x=

in that method please? im stuck, similar questions after this in book that i need to do
9. Please tell me this isn't edexcel, i've never seen anything like this on all the P3 papers i've done. Variable accleration, speed-time graphs etc come up in Mechanics but i'v never had to apply M2 to P3.. weird.
10. negative this is OCR P3 - but it should br totally unrelated you should be able to answer this with pure 3 knowledge only so im told. Also I do s2 not m2.
11. (Original post by BlueAngel)
when distance = 0 then speed = 2 ms
when distance = x then speed = 2 - 1/ 400x

time = distance / speed

x / 2 - 1/400 x = 400x^2 - 2 = x^2 = 0.005

so x = square root of 0.005

(actually thats probably wrong, im not sure.)

Or dx/ ds = k(2 - 1/ 400x)

dx / 2 - 1/ 400x = k.ds

{ 1/2 - 1/ 400x .dx = { k. ds

= LN 400X = ks

time = e^ 400x - ks + c
= e^ A400x

Then I think you have to find what C is and what A is by putting the following in:
when distance = 0 then speed = 2 ms
when distance = x then speed = 2 - 1/ 400x

Please tell me if im right or wrong. Ive got P3 next week too and Im dreading it.
12. v = 2 - (x/400)

dx/dt = 2 - x/400

dx/dt = (800 - x)/400

400int{1/(800 -x)} dx = int{t} dt

u = 800 -x
du = -dx

-400int{1/u}du = t + C

-400.ln(800 - x) = t + C

when x = 0 t = 0

C = -400.ln(800) = -2673.8

-400.ln(800 - x) = t -2673.8

so when x = 500

-400.ln(300) + 2673.8 = t

t = 392.287 seconds
= 6.53 minutes.

13. ive never come across du = -dx can elaborate on that please barrucuda?
14. i'm sorry this is just me skipping a step through habit.

u = 800 -x
du/dx = -1

and

400int{1/(u)} dx

in order to be integrating with respect to u, you need to remove the dx.

since du/dx = -1

dx can be considered -1.du

and so replaced:

-400int{1/u}du

and you can then integrate wrt u.

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