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Motion/Time Graphs

Would I be right to say that the gradient of an (t²,s) graph is acceleration, and the integral with respect to t² on an (t²,a) graph is displacement?

(Not sure whether to put this as Sixth Form or undergraduate.)

Reply 1

Brister

Original post by Big-Daddy

If by

s

you mean displacement, then that is not necessarily the case. The equation for displacement with constant acceleration, for example, is

s = ut + \frac{1}{2}at^2

. In this case, the gradient of a

(t^2,s)

graph will only be proportional to acceleration if

u = 0

.

However, acceleration obviously does not have to be constant, so the displacement equation will not always take this form anyway.

Reply 2

Big-Daddy

Original post by Brister

If by

s

you mean displacement, then that is not necessarily the case. The equation for displacement with constant acceleration, for example, is

s = ut + \frac{1}{2}at^2

. In this case, the gradient of a

(t^2,s)

graph will only be proportional to acceleration if

u = 0

.

However, acceleration obviously does not have to be constant, so the displacement equation will not always take this form anyway.

So what is the actual physical quantity represented by the gradient on a

(t^2,s)

graph, or the integral of a

(t^2,a)

graph? If there's none, we wouldn't expect it to be related to acceleration at all, would we?

Reply 3

Brister

Original post by Big-Daddy

So what is the actual physical quantity represented by the gradient on a

(t^2,s)

graph, or the integral of a

(t^2,a)

graph? If there's none, we wouldn't expect it to be related to acceleration at all, would we?

The gradient/integral will depend on the specific

s

and

a

for that situation.

By definition, though,

\frac{ds}{dt}

will always equal

v

and

\frac{dv}{dt}

will always equal

a

. I don't think

t^2

is very interesting.

Reply 4

Big-Daddy

Original post by Brister

The gradient/integral will depend on the specific

s

and

a

for that situation.

By definition, though,

\frac{ds}{dt}

will always equal

v

and

\frac{dv}{dt}

will always equal

a

. I don't think

t^2

is very interesting.

OK, if you say so - my physics textbook dropped in a free fall graph, with constant acceleration,

(t^2,s)

, and this had a constant gradient, so I was wondering if maybe acceleration was represented by the gradient of that line.

Reply 5

Brister

Original post by Big-Daddy

OK, if you say so - my physics textbook dropped in a free fall graph, with constant acceleration,

(t^2,s)

, and this had a constant gradient, so I was wondering if maybe acceleration was represented by the gradient of that line.

Yeah, it is a pretty standard experimental graph. You drop an object from rest and you will indeed get

s = \frac{1}{2}gt^2

which has a constant gradient that is used to calculate a value for the constant g. That is essentially what I was getting at in my first reply.

Reply 6

Username_valid

Original post by Big-Daddy

If the initial velocity is zero (for example if the object was released from rest), then a graph of 2s against t² will have acceleration as the gradient.