graph relationships (1 Viewer)

[Arithela]

the shape of a distance vs. time curve is determined by the velocity vs. time relationship.

why is that:

when the v/t relationship is linear then d/t is parabolic? how do you know whether its concave up or down?

when v/t is constant, d/t is linear?

 

[dolbinau]

If v/t is linear, i.e the velocity is increasing or decreasing then d/t is curved because as velocity changes the time it takes to cover the same amount of distance changes

when v/t is constant, d/t is linear because velocity remains the same so the time it takes to cover the same amount of distance does not change.

I'm not sure if it actually parabolic though, in fact unless in the first case the velocity changes directions it shouldn't be.

 

[Azreil]

d/t = integral of v/t

Say v/t = x
d/t = x^2/2
ie if v/t is x^1, d/t must be x^2, therefore parabolic.
Concave up or down is determined whether the coefficient of x^2 is positive of negative. If it's negative then it's concave down. If it's positive it's concave up.

V/t constant means v/t = k
d/t = kx + c [linear]
ie if v/t = x^0 (constant), then d/t = x^1 (linear).

 

[namburger]

when the v/t relationship is linear then d/t is parabolic?

The gradient of the curve of a distance vs time is given by the curve of the velocity vs time relationship. So if a v/t is linear, [IE: v = at + b], to find the d/t relationship, you integrate it, therefore x = at^2/2 + bt + c, where c is a constant. this is always parabolic.

how do you know whether its concave up or down

Look at a v/t linear curve, when the velocity is negative, the gradient of the x/t curve is negative and when the velocity is positive, the gradient is positive.
FOR EXAMPLE v = 2t - 1
To the left of t = 0.5, the velocity is negative and hence the gradient of the x/t graph is negative
To the right of t = 0.5, the velocity is positive and hence the gradient of the x/t graph is positive
This corresponds to a concave up parabola.

and of course, concave down is the opposite, an example would be v = -2t + 1

when v/t is constant, d/t is linear?

v = a
Integrate both sides, x = at + c, where c is a constant
this is always linear

 

[Arithela]

thanks all, but namburger you're the man!

 

[Arithela]

can anyone explain the relationship between v/t and a/t curves?

i know to get from v/t to a/t you need to differentiate but

e.g. one part of the v/t graph is parabolic (decreasing) i.e. y = x² and differentiating this gives 2x which is linear. Now, why does the answer give a curve graph rather than a linear graph for a/t during this time?

 

[Js^-1]

If you have a velocity vs time graph that is parabolic,
i.e. v=t²
Then the acceleration vs time graph is the gradient of the previous graph, since acceleration is change in velocity over change in time, i.e. dv/dt
Hence if v = t²
a = dv/dt = 2t
which is linear. Therefore if the velocity graph is parabolic, the acceleration graph will be linear. Maybe the question is asking for the Displacement vs time graph?

 

[Arithela]

hey thanks i understand all that but just dont know why the answers got it as curvy rather than linear... and yes its a a/t graph not d/t

 

[Arithela]

the question is q7 biii from 2005 paper

 

[Js^-1]

Ok, so basically, Acceleration is the gradient of the velocity vs time curve. Looking at the curve, we can see the gradient from t=0 to t=1 is 0 right, coz its a straight line across. Then we see that the curve goes downwards, which means it has negative gradient. Then at t=3, the gradient is horizontal again, so the a/t graph will be zero. From t=5 to t=6 the gradient is positive, so the acceleration graph will be positive.

Note at t=2 there is a point of inflection on the v/t graph. This translates to a maximum or minimum in the acceleration vs time graph. (Min in this case)

I will try and upload a picture, but it might be a bit difficult.

 

[Js^-1]

Hope this helps...

 

[Arithela]

ah.. so the point of inflexion means there won't be two linear lines joining between t=1 and t=3 but instead a parabolic shape?

 

[Captain Hero]

This dude I know married a box plot. They're pretty happy, I guess. I don't think it'll work out in the long term.

 

[Js^-1]

ah.. so the point of inflexion means there won't be two linear lines joining between t=1 and t=3 but instead a parabolic shape?

Yeh...If there is a point of inflection, then the equation between those two points is at least of degree 3 (a cubic or more). Since it has a degree of
> or = 3
then its derivative, the acceleration vs time graph, will have a degree of
> or = 2.
So the acceleration vs time graph shouldn't really be linear between those two points.

 

[Arithela]

thank u so much! ill keep that in mind

 

[Js^-1]

No problem, glad I could help.