MJ20 P12 Q6 Car Distance Graph | May/June 2020 | CAIE A Level 9702 Physics

9702/12/M/J/20: A car X is travelling at a constant speed u along a straight road. At time t = 0 a second car Y is a distance d0 behind car X and travelling at a speed v in the same direction. Speed v is less than speed u.
At time t = 0 car Y begins to accelerate with a constant acceleration. Car Y overtakes car X at time t = T. Which graph could best show the variation with time t of the distance d between the cars?
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[@HabibUrRahman-sr2wn]

Your way of explaination is very satisfactory .... you will be definitely a very good teacher in physical classes

[@sierqhan325]

You mentioned it will not be linear due to s =t² tgrough s=ut +1/2at² but according to s= 1/2 (u+v)t the relationship isnt the same

[@layanodah5151]

I think I'm late and I'm not sure if this will help, but in s= 1/2 (u+v)t both v and t are changing, so you can't see how s changes relative to t only. In the first equation, u and a are both constant, so you can see the direct variation of s with t. When writing a formula to see how two variables vary relative to eachother, it's important that all other variables are constant. (Like for example EPE= ½Fx and EPE= ½kx², if F and x are both changing, but you want to see how EPE varies with x, you'll have to go for the second formula as k is constant).

[@sierqhan325]

@@layanodah5151 p1 in a few hours, this was very helpful ❤

[@layanodah5151]

​@@sierqhan325 I'm glad this helped! I hope the paper goes well today

[@soha6734]

please help i still didn't get the last thing, why does the graph in B increases first and then decreases why can't it be C

[@justshan2413]

This was really helpful.

[@AA-qg2hu]

How would I calculate how fast car y is traveling when it passes car x? Please help!

[@Heyjudeeeee12321]

think this way using concept of relative velocity .. see the car X is moving at a constant velocity so its velocity remains same throughout the journey . BUT the car Y is accelerating now ..think of numerical values closer to them let us give the car Y a speed of 5 ms-1 and the car X 8 ms-1 . the both cars are moving at the same direction . since the car Y is accelerating it will meet car X at some point. so the point or the part of journey they meet WITH SAME VELOCITIES ANF the relative velocity will be 0 . when car X and car Y are at the same point they would observe that neither of them is moving . now look at the d/t graphs at graph B and u can draw a gradient on the and you will see the velocity is 0 AT time T none of the other graph has that

[@nytman3192]

why do we use a SUVAT eq to relate dist and time? why not the good old v=d/t, that way we know its a linear eq and its going to be a straight line? Plz help, got my CIE tomorrow

[@ETphysics]

v=d/t comes from a SUVAT equation, but it assumes acceleration is zero. s = ut + ½at² s = ut + 0 Since v and u are the same because no acceleration, they are interchangeable d = vt v = d/t

[@PHYSICS_ACE_GCSE_IGCSE_A-LEVEL]

I need a help.