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Can someone please help me with a misunderstanding in equations of motions,

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?

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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In addition to that can someone please answer this question

9702_s17_qp_13, question 7, the answer is D

Thanks in advance

9702_s17_qp_13, question 7, the answer is D

Thanks in advance

Last edited by ZedEmDoubleU; 2 years ago

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#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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#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 ****

I think. idk i might be chatting ****

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#5

(Original post by

Can someone please help me with a misunderstanding in equations of motions,

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?

**ZedEmDoubleU**)Can someone please help me with a misunderstanding in equations of motions,

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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#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.

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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#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

Can someone please help me with a misunderstanding in equations of motions,

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?

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.

**ZedEmDoubleU**)

Can someone please help me with a misunderstanding in equations of motions,

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

reply

ohhh right

thx a lot

thx a lot

(Original post by

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.

**mohamadjamil03**)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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so based on the variables im given, i use one or the other?

(Original post by

s and t are not constant, and you're ignoring the offsets.

**RogerOxon**)s and t are not constant, and you're ignoring the offsets.

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#10

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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can u explain what you mean by "derivative or integral of one other" ?

(Original post by

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.

**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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#12

(Original post by

can u explain what you mean by "derivative or integral of one other" ?

**ZedEmDoubleU**)can u explain what you mean by "derivative or integral of one other" ?

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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can someone attempt this, and explain please,

Thank you ALL for all the help you've provided

Thank you ALL for all the help you've provided

(Original post by

In addition to that can someone please answer this question

9702_s17_qp_13, question 7, the answer is D

Thanks in advance

**ZedEmDoubleU**)In addition to that can someone please answer this question

9702_s17_qp_13, question 7, the answer is D

Thanks in advance

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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

yeah thats a great idea, i'll try it thanks

(Original post by

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.

**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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#15

(Original post by

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

**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

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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#16

^All of that is because of F=ma. Force ∝ acceleration provided the mass doesn’t change

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(Original post by

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.

**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.

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