# How to calculate acceleration on a distance time graph?

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Megan_101

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#1

Indeterminate

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JRM3PM

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Thank you!

**Megan_101**)Thank you!

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Arbolus

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#4

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.

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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#5

(Original post by

It depends. What's the question?

**Indeterminate**)It depends. What's the question?

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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#6

(Original post by

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.

**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.

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#7

(Original post by

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.

**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.

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z33

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(Original post by

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.

**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.

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JRM3PM

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(Original post by

It is a curved line.

**Megan_101**)It is a curved line.

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#10

(Original post by

draw a tangent at that point?

**z33**)draw a tangent at that point?

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z33

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#11

(Original post by

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?

**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?

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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#12

(Original post by

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

**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 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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#13

(Original post by

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?

**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?

so yeah that should be correct

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#14

(Original post by

yeah if the object starts from rest then u is 0

so yeah that should be correct

**z33**)yeah if the object starts from rest then u is 0

so yeah that should be correct

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#15

(Original post by

I'm very grateful for your help and time

**Megan_101**)I'm very grateful for your help and time

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lerjj

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#16

**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

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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#17

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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#18

How can you take the gradient as velocity? It is a distance time graph, not a displacement time graph?

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