How To Find Distance In Acceleration Time Graph

Information technology depends. What'south the question?

Can't be washed easily. Acceleration/deceleration on a distance time graph would be seen as a change of gradient in a line. AKA a curved line.

2d derivative of deportation with respect to fourth dimension.

i.e. have the slope of the graph to find the velocity at whatsoever given moment, and then accept the gradient of that to notice what the acceleration is. If the altitude-time graph is a straight line, then the acceleration in the direction of motion is zero.

(Original mail service by Indeterminate)
Information technology depends. What's the question?

Office a) Employ the distance time graph to determine the speed of the object at a fourth dimension of 4s - which it irked outs I be 4ms-one
Office b) literally all it says is "calculate the acceleration". I'm assuming it somehow relates to the first part of the question only I accept no clue how- havent been taught before.

Thank you lot for your respond.

(Original post by york_wbu)
Tin 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.

(Original mail past Arbolus)
Second derivative of displacement with respect to time.

i.due east. take the slope of the graph to find the velocity at any given moment, and so take the slope of that to find what the acceleration is. If the distance-fourth dimension graph is a straight line, and so the acceleration in the direction of movement is zero.

Okay so the velocity at a given indicate is 4ms-i. How am I meant to accept the gradient of that? The line is curved past the way.

(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 manner.

depict a tangent at that point?

If the line is curving at a constant charge per unit, I guess y'all could utilise two tangents. i.e find the speed at ii 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/fourth dimension.

Cheers. I've done that and it immune me I get a another signal that the tangent crosses (three,viii). Past doing this I calculated the gradient betwixt the points (iii,eight) and (4,xvi) which was 8. Does this mean the dispatch would be 4ms-1/ the gradient of the line which was eight?

(Original mail service by Megan_101)
Thank you lot. I've done that and it allowed me I go a another point that the tangent crosses (iii,8). Past doing this I calculated the gradient betwixt the points (3,8) and (four,16) which was viii. Does this mean the dispatch would exist 4ms-1/ the slope of the line which was 8?

it's a curve right? doesnt that mean you need to have 2 unlike tangents?

then dispatch = (v-u)/t

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

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

think thats right^ been a long time since i did this and then double bank check the answer just i recall thats how y'all do it

(Original postal service by z33)
information technology'south a bend right? doesnt that mean you demand to take 2 different tangents?

and then dispatch = (5-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 ii = 5
and the fourth dimension is on the x centrality, just count information technology between the 2 points y'all drew the tangents at

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

remember thats right^ been a long time since i did this and so double cheque the answer only i recall thats how y'all practice it

For the initial velocity (tangent ane) could I use 0? The graph passes through the origin so the velocity would exist 0 at that point?
If so,

Gradient ii (8) - gradient1 (0) / change in time (4s) so the acceleration is 2ms-2?

(Original post by Megan_101)
For the initial velocity (tangent 1) could I use 0? The graph passes through the origin and so the velocity would exist 0 at that point?
If so,

Slope ii (8) - gradient1 (0) / change in time (4s) then the acceleration is 2ms-ii?

yeah if the object starts from rest then u is 0

so yeah that should be correct

(Original post by z33)
yes if the object starts from remainder so u is 0

and then yeah that should be correct

I'g very grateful for your assistance and time

(Original postal service past Megan_101)
I'g very grateful for your aid and fourth dimension

it'south fine don't worry about it

(Original post by z33)
it's a curve right? doesnt that hateful you need to have 2 different tangents?

and then dispatch = (five-u)/t

the gradient of a distance fourth dimension 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 centrality, just count it between the 2 points you drew the tangents at

then substitute: (gradient2 - gradient1)/ change in fourth dimension

recollect thats correct^ been a long time since i did this so double check the answer but i remember thats how you do it

If the acceleration is constant (i.due east. if it's a parabola or x^two graph) then that will always piece of work. Otherwise, this will still be pretty good then long as you put the two tangents quite well-nigh the bespeak of interest, and ideally have them arranged symmetrically. I.e. take one tangent to find the velocity 1s earlier the the fourth dimension you want the acceleration for, and another one 1s afterwards, and so use the formula you gave.

Just yes, I tin't see whatever way of measuring the 2nd derivative directly other than past drawing multiple tangents and doing some kind of calculation using their gradients.

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 slope from the first to the second is the same as the change in slope from the second to the third, you tin can exist reasonably confident that the acceleration is constant and that your (two) measurement(s) of it are accurate.

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