velocity help

#1
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?
0
7 months ago
#2
(Original post by dwrfwrw)
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?
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.
Last edited by mqb2766; 7 months ago
0
7 months ago
#3
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.
0
#4
(Original post by 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.
Thanks so much!!

(Original post by mqb2766)
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.
thank you so much very helpful
1
7 months ago
#5
(Original post by dwrfwrw)
Thanks so much!!

thank you so much very helpful
Happy to help
0
#6
(Original post by mqb2766)
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.
through relative motion, do you mean suvat since you told me that v-t graphs were paramatrized forms of constant acceleration formulae?
0
7 months ago
#7
(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.
Last edited by mqb2766; 7 months ago
0
#8
(Original post by mqb2766)
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.
never thought of it that way
0
7 months ago
#9
(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.
0
#10
(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?
0
7 months ago
#11
(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.
0
#12
(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.
Oh okay, makes sense thanks so much I fully understand now
0
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