#1

I have a physics test tomorrow and i was hoping the pit could help me with this one physics concept.

When you are trying to find the instantaneous velocity of something, you are supposed to find the tangent line to that point on the velocity graph. I dont understand how to find the tangent line.

When you are trying to find the instantaneous velocity of something, you are supposed to find the tangent line to that point on the velocity graph. I dont understand how to find the tangent line.

#2

You like Wintersun, so I feel bad as I have no idea and can't help you

#3

I hate drawing tangents. You're supposed to move a ruler closer to the curve until it just touches the point at which you are measuring the velocity. Then draw the line.

But if you're talking about a velocity/time graph then drawing a tangent is pointless.

But if you're talking about a velocity/time graph then drawing a tangent is pointless.

*Last edited by EuBoat at Sep 21, 2009,*

#4

Have you taken any calculus classes? If so just take the derivative of the equation representing velocity.

#5

I hate drawing tangents. You're supposed to move a ruler closer to the curve until it just touches the point at which you are measuring the velocity. Then draw the line.

But if you're talking about a veliovity/time graph then drawing a tangent is pointless.

it says something about using the x-intercept of the tangent line though ...

#6

I have a physics test tomorrow, too, but it's over totally different topics

#7

Shouldn't you just look at the velocity on the y axis of the graph?

I'm so sure you're not meant to draw a tangent for a static value.

I'm so sure you're not meant to draw a tangent for a static value.

#8

depends on the graph.

if its velocity/time grpah, just find the time and the point on the curve is the instantaneous velocity

if its displacement/time, you need the gradient of the tangent. the tangent is a straight line which touches the curve only at one point. draw it, then you meause the gradient by dividing the change in the Y-direction of the tangent by the change in the X-direction on the tangent. As in, take two co-ordinates on the tangent.

ummm i think thats all you need, unless they give you an acceleration/time graph, in whcih case you need to integrate it. But i doubt they will considering you dont understand about displacement/time ones yet.

hope this helps

if its velocity/time grpah, just find the time and the point on the curve is the instantaneous velocity

if its displacement/time, you need the gradient of the tangent. the tangent is a straight line which touches the curve only at one point. draw it, then you meause the gradient by dividing the change in the Y-direction of the tangent by the change in the X-direction on the tangent. As in, take two co-ordinates on the tangent.

ummm i think thats all you need, unless they give you an acceleration/time graph, in whcih case you need to integrate it. But i doubt they will considering you dont understand about displacement/time ones yet.

hope this helps

#9

Aha, I just did this in Advanced Functions. The proper way to do is to find the slope of a secant that passes through the actual point and a point very close to it (about 0.001 seconds before or after).

#10

can you post the question?

#11

depends on the graph.

if its velocity/time grpah, just find the time and the point on the curve is the instantaneous velocity

if its displacement/time, you need the gradient of the tangent. the tangent is a straight line which touches the curve only at one point. draw it, then you meause the gradient by dividing the change in the Y-direction of the tangent by the change in the X-direction on the tangent. As in, take two co-ordinates on the tangent.

ummm i think thats all you need, unless they give you an acceleration/time graph, in whcih case you need to integrate it. But i doubt they will considering you dont understand about displacement/time ones yet.

hope this helps

oh ok thanks i think i understand it now

#12

Well I haven't looked at v/t graphs for a while so I'm a little rusty I suppose. But if (on a v/t graph) you wanted to find the velocity at a given time you'd just look at the the y-value for velocity corresponding to that time (x-value)it says something about using the x-intercept of the tangent line though ...

#13

oh ok thanks i think i understand it now

forgot to mention, though you probs guessed, when you draw the tangent, the point it touches the curve should be where the time value is where u need the instantaneous velocity

think thats all... =)

#14

if its a velocity time graph all you have to do is look at the y- value (i did all this stuff 2 years ago so its a little fuzzy)

#15

I would never have expected THE PIT of all places to answer this seriously. I was ready to hear a bunch of funny jokes.

#16

if its displacement/time, you need the gradient of the tangent. the tangent is a straight line which touches the curve only at one point. draw it, then you meause the gradient by dividing the change in the Y-direction of the tangent by the change in the X-direction on the tangent. As in, take two co-ordinates on the tangent.

Dont do this, its way way too inaccurate. Instead, do this:

Aha, I just did this in Advanced Functions. The proper way to do is to find the slope of a secant that passes through the actual point and a point very close to it (about 0.001 seconds before or after).

This means doing a normal slope calculation, where delta-y/delta-x= slope. Ie, if you need the velocity at t=2.5, do y2-y1/2.50001-2.50.

The other time should just be very close to the time needed.

#17

Have you taken any calculus classes?If so just take the derivative of the equation representing velocity.

THIS. That's all you need to do.