How to calculate acceleration on a distance time graph? Watch

Megan_101
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Thank you!
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Indeterminate
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(Original post by Megan_101)
Thank you!
It depends. What's the question?
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JRM3PM
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(Original post by Megan_101)
Thank you!
Can't be done easily. Acceleration/deceleration on a distance time graph would be seen as a change of gradient in a line. AKA a curved line.
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Arbolus
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Second derivative of displacement with respect to time.

i.e. take the gradient of the graph to find the velocity at any given moment, and then take the gradient of that to find what the acceleration is. If the distance-time graph is a straight line, then the acceleration in the direction of motion is zero.
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Megan_101
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(Original post by Indeterminate)
It depends. What's the question?
Part a) Use the distance time graph to determine the speed of the object at a time of 4s - which it irked outs I be 4ms-1
Part b) literally all it says is "calculate the acceleration". I'm assuming it somehow relates to the first part of the question but I have no clue how- havent been taught before.

Thank you for your reply.
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Megan_101
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(Original post by york_wbu)
Can't be done easily. Acceleration/deceleration on a distance time graph would be seen as a change of gradient in a line. AKA a curved line.
It is a curved line.
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Megan_101
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(Original post by Arbolus)
Second derivative of displacement with respect to time.

i.e. take the gradient of the graph to find the velocity at any given moment, and then take the gradient of that to find what the acceleration is. If the distance-time graph is a straight line, then the acceleration in the direction of motion is zero.
Okay so the velocity at a given point is 4ms-1. How am I meant to take the gradient of that? The line is curved by the way.
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z33
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(Original post by Megan_101)
Okay so the velocity at a given point is 4ms-1. How am I meant to take the gradient of that? The line is curved by the way.
draw a tangent at that point?
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JRM3PM
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(Original post by Megan_101)
It is a curved line.
If the line is curving at a constant rate, I guess you could use two tangents. i.e find the speed at two separate points along the curve. Then use these speeds and the difference in these points along the time axis to work out the change in speed/time.
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Megan_101
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(Original post by z33)
draw a tangent at that point?
Thank you. I've done that and it allowed me I get a another point that the tangent crosses (3,8). By doing this I calculated the gradient between the points (3,8) and (4,16) which was 8. Does this mean the acceleration would be 4ms-1/ the gradient of the line which was 8?
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z33
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(Original post by Megan_101)
Thank you. I've done that and it allowed me I get a another point that the tangent crosses (3,8). By doing this I calculated the gradient between the points (3,8) and (4,16) which was 8. Does this mean the acceleration would be 4ms-1/ the gradient of the line which was 8?
it's a curve right? doesnt that mean you need to have 2 different tangents?

so acceleration = (v-u)/t

the gradient of a distance time graph is the speed
so calculate the gradient of tangent 1 = u
and the gradient of tangent 2 = v
and the time is on the x axis, just count it between the 2 points you drew the tangents at

then substitute: (gradient2 - gradient1)/ change in time

think thats correct^ been a long time since i did this so double check the answer but i think thats how you do it
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Megan_101
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(Original post by z33)
it's a curve right? doesnt that mean you need to have 2 different tangents?

so acceleration = (v-u)/t

the gradient of a distance time graph is the speed
so calculate the gradient of tangent 1 = u
and the gradient of tangent 2 = v
and the time is on the x axis, just count it between the 2 points you drew the tangents at

then substitute: (gradient2 - gradient1)/ change in time

think thats correct^ been a long time since i did this so double check the answer but i think thats how you do it
For the initial velocity (tangent 1) could I use 0? The graph passes through the origin so the velocity would be 0 at that point?
If so,

Gradient 2 (8) - gradient1 (0) / change in time (4s) so the acceleration is 2ms-2?
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z33
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(Original post by Megan_101)
For the initial velocity (tangent 1) could I use 0? The graph passes through the origin so the velocity would be 0 at that point?
If so,

Gradient 2 (8) - gradient1 (0) / change in time (4s) so the acceleration is 2ms-2?
yeah if the object starts from rest then u is 0

so yeah that should be correct
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Megan_101
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(Original post by z33)
yeah if the object starts from rest then u is 0

so yeah that should be correct
I'm very grateful for your help and time
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z33
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(Original post by Megan_101)
I'm very grateful for your help and time
it's fine don't worry about it
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lerjj
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(Original post by z33)
it's a curve right? doesnt that mean you need to have 2 different tangents?

so acceleration = (v-u)/t

the gradient of a distance time graph is the speed
so calculate the gradient of tangent 1 = u
and the gradient of tangent 2 = v
and the time is on the x axis, just count it between the 2 points you drew the tangents at

then substitute: (gradient2 - gradient1)/ change in time

think thats correct^ been a long time since i did this so double check the answer but i think thats how you do it
If the acceleration is constant (i.e. if it's a parabola or x^2 graph) then that will always work. Otherwise, this will still be pretty good so long as you put the two tangents quite near the point of interest, and ideally have them arranged symmetrically. I.e. take one tangent to find the velocity 1s before the the time you want the acceleration for, and another one 1s afterwards, then use the formula you gave.

But yes, I can't see any way of measuring the 2nd derivative directly other than by drawing multiple tangents and doing some kind of calculation using their gradients.
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mik1a
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To expand on other (correct) answers, a real world solution would be to measure three tangents at 1 second intervals (or some other sensible time unit from the graph). If the change in gradient from the first to the second is the same as the change in gradient from the second to the third, you can be reasonably confident that the acceleration is constant and that your (two) measurement(s) of it are accurate.
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Ibra007
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How can you take the gradient as velocity? It is a distance time graph, not a displacement time graph?
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