# Physics velocity question

Hi, please could i have help on this question? I thought the answer should be B because it is uniform deceleration and constant velocity but the answer is A?
Here is the question: https://app.gemoo.com/share/image-annotation/582940842854494208?codeId=PYxNX28pdRJgO&origin=imageurlgenerator
Thank you very much!
Be careful - the axes are not velocity versus time, they are velocity versus distance (also remember that the velocity will not be constant once brakes are applied).
Original post by lordaxil
Be careful - the axes are not velocity versus time, they are velocity versus distance (also remember that the velocity will not be constant once brakes are applied).

Thank you for your reply. It says uniform deceleration so does that not mean that it will still be a straight line because it is constant, how would the x axis being distance change that? Also, it says uniform deceleration after the brakes are applied so would it not be constant? Thank you!
If the car is decelerating uniformly (after brakes are applied) then the velocity will decrease linearly with time. This is why you may have thought B was the right answer. However, the x-axis shows distance, not time, and the velocity will not decrease linearly with distance for uniform deceleration. Do you know the SUVAT formulae?
Original post by anonymous294
Thank you for your reply. It says uniform deceleration so does that not mean that it will still be a straight line because it is constant, how would the x axis being distance change that? Also, it says uniform deceleration after the brakes are applied so would it not be constant? Thank you!

Think about the distances covered in going from 70->65 mph as opposed to 5->0 mph. Would they be the same distance (with the same braking force) or ...?

The relevant suvat is the one not involving time so,
v^2 = u^2 + 2as
So its a quadratic relationship between v (y axis) and s (x axis) as a and u are constant so make sure you can square that with the graphs/your intuition.
(edited 3 months ago)
Original post by mqb2766
Think about the distances covered in going from 70->65 mph as opposed to 5->0 mph. Would they be the same distance (with the same braking force) or ...?

The relevant suvat is the one not involving time so,
v^2 = u^2 + 2as
So its a quadratic relationship between v (y axis) and s (x axis) as a and u are constant so make sure you can square that with the graphs/your intuition.

For your first question, would the distance be the same because the change in velocity for both is the same so when it is squared it would still be equal and you are dividing by the constant ‘a’ so s will be equal?
Also, I sort of understand what you mean by the relations between v and s. Does the formula show that the graph will be s and v^2 so instead of being a straight line for acceleration, it will be squared so turn into a curve instead?
Thank you very much!
Original post by anonymous294
For your first question, would the distance be the same because the change in velocity for both is the same so when it is squared it would still be equal and you are dividing by the constant ‘a’ so s will be equal?
Also, I sort of understand what you mean by the relations between v and s. Does the formula show that the graph will be s and v^2 so instead of being a straight line for acceleration, it will be squared so turn into a curve instead?
Thank you very much!

Youre in a car and when you brake with the same force youre saying that you would travel the same distance going from 70->65 and 5->0? The first one is motorway braking so you tavel a long way, the second youd travel a gnats whisker. Surely?

Youre hung up on the x-axis being time again. If the time is the same (with constant acceleration) for the same change in speed, you must travel further in the first case as the average speed during braking is (much) larger.

The (suvat parabolic) curve is
y^2 = c + mx
where m is negative and c is positive. Can you sketch it, so what is the maximum value of x (what does it represent in terms of braking), ...
(edited 3 months ago)
Original post by mqb2766
Youre in a car and when you brake with the same force youre saying that you would travel the same distance going from 70->65 and 5->0? The first one is motorway braking so you tavel a long way, the second youd travel a gnats whisker. Surely?

Youre hung up on the x-axis being time again. If the time is the same (with constant acceleration) for the same change in speed, you must travel further in the first case as the average speed during braking is (much) larger.

The (suvat parabolic) curve is
y^2 = c + mx
where m is negative and c is positive. Can you sketch it, so what is the maximum value of x (what does it represent in terms of braking), ...

Hi, thank you for your reply. I’m still quite lost, I understand the first two parts where you would travel a larger distance in the first case but I don’t understand how the curves is got to do with this and how I can sketch it becuase mx+c is linear and y^2 is quadratic? Thank you
Original post by anonymous294
Hi, thank you for your reply. I’m still quite lost, I understand the first two parts where you would travel a larger distance in the first case but I don’t understand how the curves is got to do with this and how I can sketch it becuase mx+c is linear and y^2 is quadratic? Thank you

Id really work it through for a simple example (for both the calculation and sketch). So how far would you travel in the two cases

1.

10->9 m/s with a deceleration of 1m/s^2

2.

1->0 m/s with a deceleration of 1m/s^2

So what can you deduce in general / what is the shape of the dispacement/velocity graph?

For sketching the parabola for v^2=u^2+2as with a=-1 and u=10 so
y^2 = 100 - 2x
Can you stick a few numbers in (for x (s)) and plot the y (v), can you relate it to a normal form, .. would it be easier to think about if it was
x^2 = 100 - 2y
or equivalently
y = -1/2 x^2 + 50
and how are the two related, ...?

y = x^2
y^2 = x or y = +/-sqrt(x)
and think about inverse function / reflection in the line y=x and how you can make the parabolas a bit more general. At the end of the day, constant acceleleration suvats are simply linear/parabolic functions which are covered to death in maths gcse, but its easy to miss it when you come to apply them.

Best advice is to upload what youve tried, especially if youre having a bit of a block with the question(s) and get in the habit of setting up / thinking about simple examples.
(edited 3 months ago)
Original post by anonymous294
Hi, please could i have help on this question? I thought the answer should be B because it is uniform deceleration and constant velocity but the answer is A?
Here is the question: https://app.gemoo.com/share/image-annotation/582940842854494208?codeId=PYxNX28pdRJgO&origin=imageurlgenerator
Thank you very much!

You would be correct
B
If the horizontal axis is TIME
The gradient of the sloping line is constant and negative so it would be B
BUT
the horizontal axis is DISTANCE

This is hard for GCSE
Look at picture
Numerator is acceleration constant
Denominator is velocity which is REDUCING

(edited 3 months ago)