# Physics A level question

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#1
one of the equations is v^2 = u^2 + 2as meaning that v^2 ∝ a
but another equation, v = u + at, implies that v ∝ a
which one is correct?
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#2
9702_s17_qp_13, question 7, the answer is D
Last edited by ZedEmDoubleU; 2 years ago
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2 years ago
#3
They’re both correct. Proportionality isn’t the same as an equals sign. Both v and v² are proportional to a.
Last edited by hi_imcatherine; 2 years ago
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2 years ago
#4
neither of your assumptions are correct. v^2 proportional to a (plus u^2) and v is proportional to a (plus u)
I think. idk i might be chatting ****
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2 years ago
#5
(Original post by ZedEmDoubleU)
one of the equations is v^2 = u^2 + 2as meaning that v^2 ∝ a
but another equation, v = u + at, implies that v ∝ a
which one is correct?
s and t are not constant, and you're ignoring the offsets.
1
2 years ago
#6
I think it’s important to consider the fact that these proportionalities only hold in the context of their respective equations. And outside of suvat / uniform acceleration, you shouldn’t be thinking of their proportionality, but rather their relationship as derivative & integral of one another.

So v ∝ a when u is added, and when a is being multiplied by t. But as you spotted, there are different relationships between these variables in different equations.
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2 years ago
#7
Neither of those proportionality equations look right. Usually, when we talk about proportionality, we say the thing on the left is proportional to something on the right IF there is only one term on the right. Here's an example:

distance = speed * time, therefore distance is directly proportional to speed and distance is directly proportional to time.

However, as soon as I add something to the right side e.g. in 'v = u + at', you can't just say that v is proportional to a. Instead, you can subtract u from both sides and say something like: (v - u) is proportional to a.
(Original post by ZedEmDoubleU)
one of the equations is v^2 = u^2 + 2as meaning that v^2 ∝ a
but another equation, v = u + at, implies that v ∝ a
which one is correct?
1
#8
ohhh right
thx a lot

Neither of those proportionality equations look right. Usually, when we talk about proportionality, we say the thing on the left is proportional to something on the right IF there is only one term on the right. Here's an example:

distance = speed * time, therefore distance is directly proportional to speed and distance is directly proportional to time.

However, as soon as I add something to the right side e.g. in 'v = u + at', you can't just say that v is proportional to a. Instead, you can subtract u from both sides and say something like: (v - u) is proportional to a.
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#9
so based on the variables im given, i use one or the other?

(Original post by RogerOxon)
s and t are not constant, and you're ignoring the offsets.
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2 years ago
#10
(Original post by ZedEmDoubleU)
so based on the variables im given, i use one or the other?
Yes.

The variables are not independent - if you vary a, with constant u and t, v and s must change, etc.
Last edited by RogerOxon; 2 years ago
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#11
can u explain what you mean by "derivative or integral of one other" ?

(Original post by hi_imcatherine)
I think it’s important to consider the fact that these proportionalities only hold in the context of their respective equations. And outside of suvat / uniform acceleration, you shouldn’t be thinking of their proportionality, but rather their relationship as derivative & integral of one another.

So v ∝ a when u is added, and when a is being multiplied by t. But as you spotted, there are different relationships between these variables in different equations.
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2 years ago
#12
(Original post by ZedEmDoubleU)
can u explain what you mean by "derivative or integral of one other" ?
Have you done any calculus yet? Most kids cover it in maths first, but I’ll briefly explain.

If you differentiate a graph that describes how displacement varies with time, the result is a new graph that describes the gradient of that graph. Since the gradient of a displacement-time graph is the velocity, the derivative (result of differentiation) of change in displacement with time is velocity.

You can do the same with a velocity-time graph, if you differentiate that function, you get a function of acceleration. ie. acceleration is the derivative of velocity. Integration is the opposite of differentiation, so velocity is the integral of acceleration, and displacement is the integral of velocity. These relationships are totally different to proportionality, but they’re much more useful to consider in stuff like the question you added.

I’m sorry that my first post was wrong!! It is definitely true that you can’t state proportionalities like that when there are other variables involved.
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#13
can someone attempt this, and explain please,

(Original post by ZedEmDoubleU)
9702_s17_qp_13, question 7, the answer is D
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#14
So you mean i should try the derivative or the integral to kind of find what the graph should look like?
yeah thats a great idea, i'll try it thanks

(Original post by hi_imcatherine)
Have you done any calculus yet? Most kids cover it in maths first, but I’ll briefly explain.

If you differentiate a graph that describes how displacement varies with time, the result is a new graph that describes the gradient of that graph. Since the gradient of a displacement-time graph is the velocity, the derivative (result of differentiation) of change in displacement with time is velocity.

You can do the same with a velocity-time graph, if you differentiate that function, you get a function of acceleration. ie. acceleration is the derivative of velocity. Integration is the opposite of differentiation, so velocity is the integral of acceleration, and displacement is the integral of velocity. These relationships are totally different to proportionality, but they’re much more useful to consider in stuff like the question you added.

I’m sorry that my first post was wrong!! It is definitely true that you can’t state proportionalities like that when there are other variables involved.
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2 years ago
#15
(Original post by ZedEmDoubleU)
So you mean i should try the derivative or the integral to kind of find what the graph should look like?
yeah thats a great idea, i'll try it thanks
Yes exactly. In the context of your question, you can think of the change in force as change in acceleration. And slowly increasing the braking force means slowly increasing the deceleration. You know that the acceleration/deceleration is the derivative of the speed-time graph, so you need to find the graph with a gradient that slow increases.

You can see why it’s D because the rate of change of speed (the gradient/deceleration) is slow at first, and increases with time.
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2 years ago
#16
^All of that is because of F=ma. Force ∝ acceleration provided the mass doesn’t change
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#17
(Original post by hi_imcatherine)
Yes exactly. In the context of your question, you can think of the change in force as change in acceleration. And slowly increasing the braking force means slowly increasing the deceleration. You know that the acceleration/deceleration is the derivative of the speed-time graph, so you need to find the graph with a gradient that slow increases.

You can see why it’s D because the rate of change of speed (the gradient/deceleration) is slow at first, and increases with time.
Yeaaa thx!
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