velocity help

https://gyazo.com/70f72d52c8bacee94eca79499d663bc2
I'm really confused on question c
https://gyazo.com/70f72d52c8bacee94eca79499d663bc2
I managed to draw the diagram https://gyazo.com/b82da33cea4e70aa21242217c74fe344
but I'm not really sure how to approach this question. I attempted to find areas under the graph and perhaps subtract area from above from below?

Reply 1

mqb2766

Original post by dwrfwrw

As you say work out the distances travelled at a time by finding the areas under the graph. For c) when are the areas equal? Ive put a bit more detail in your other thread.

Note, you could do it using relative motion.

(edited 2 years ago)

Reply 2

IJaguar

Using your answer to b, you know the distance between P and Q when P reaches its maximum speed. You also know that P's maximum speed is 10m/s faster than Q's maximum, and so the distance between P and Q is decreasing by 10 metres every second.
Using this information, you can divide the distance found in b by 10 to find the answer to c.

Reply 3

dwrfwrw

Original post by IJaguar

Thanks so much!!

Original post by mqb2766

thank you so much very helpful

Reply 4

IJaguar

Original post by dwrfwrw

Thanks so much!!

thank you so much very helpful

Happy to help :smile:

Reply 5

dwrfwrw

Original post by mqb2766

through relative motion, do you mean suvat since you told me that v-t graphs were paramatrized forms of constant acceleration formulae?

Reply 6

mqb2766

Original post by dwrfwrw

through relative motion, do you mean suvat since you told me that v-t graphs were paramatrized forms of constant acceleration formulae?

There are two different constant acceleration phases here, so youd have to do two suvat phases. Its possible, but its arguably easier to do the ad hoc area calculation as per the other thread. If you assumed Q was constant (so the observer is on board), then the motion of P would be described by
u = -20
a = 2
When t<=15. So youd get s(15) = -75.

Then for t greater than 15
s(0)=-75
u=10
a=0
and youd be finding the time when s=0.

(edited 2 years ago)

Reply 7

dwrfwrw

Original post by mqb2766

never thought of it that way

Reply 8

mqb2766

Original post by dwrfwrw

never thought of it that way

Sometimes it does simplify the problem to imagine the observer on one vehicle and analyse the relative motion of the other. Sometimes not. But its worth considering if you practice it a bit.

Reply 9

dwrfwrw

Original post by mqb2766

Sometimes it does simplify the problem to imagine the observer on one vehicle and analyse the relative motion of the other. Sometimes not. But its worth considering if you practice it a bit.

just as a heads up I understood your method however I'm unsure how you got u=-20 m/s, is it to make sure initial conditions are the same?

Reply 10

mqb2766

Original post by dwrfwrw

just as a heads up I understood your method however I'm unsure how you got u=-20 m/s, is it to make sure initial conditions are the same?

Initially Q is going 20 m/s faster than P, so P is going 20 m/s slower than Q.
If the observer is on Q, that means the initial relative velocity of P is -20 m/s.

Reply 11

dwrfwrw

Original post by mqb2766

Initially Q is going 20 m/s faster than P, so P is going 20 m/s slower than Q.
If the observer is on Q, that means the initial relative velocity of P is -20 m/s.