Question

if you were to calculate the area between a line and the x-axis on a velocity-time graph, what would you find. I believe it is distance traveled but not sure why

Answer 1

what class is this?

You are correct, but it would help to know the class this is in before explaining.

Answer 2

physics

Answer 3

have you done integral calculus yet?

Answer 4

The easiest way to see this without calculus is to plot a line for velocity that is a constant value... say velocity = 10 feet per second.

Then on the horizontal "time" access, calculate the area between 0 and 5 seconds... it's a rectangle with length 5 and height 10, or 50 feet traveled.

Answer 5

you could also do a sloped line for velocity, meaning velocity is increasing linearly... the shape under the curve would be a triangle, and you can find that area which will be distance traveled.

In calculus, you learn how to integrate functions, and distance traveled (as a function of time) is the integral of velocity (as a function of time).

Integrals for lines are exactly the same thing as area under the curve through geometry methods like L x W or (1/2) x base x height

Answer 6

The area between the x-axis which I can find via calculus Sdx, is either distance traveled or displacement but I am not sure which one and why

Answer 7

are you trying to distinguish between "distance traveled" and "displacement"? Sounds sort of similar to me, depending on the physical example... an object could be displaced by an amount during a period of time if it was moving at some velocity, but you could also describe that object as "traveling a distance".

I'm not sure I see the distinction, but maybe I am missing something.

Answer 8

the graphix is a straight line of a walker going acoss with a unit of 1 on the Y axis. the units are in m/s. the area under that line Ibelieve is distance traveled but not sure why

Answer 9

but you can do the integral?

if the velocity as a function of time is v(t), then in this question, v(t) = 1.... it's a constant velocity of 1 m/s.

The integral of v(t) dt is t, then evaluated over the t values in question.

So, for the region t = 0 to t = 2,
the integral of v(t)dt = t evaluates to 2 - 0 = 2 meters.

It's identical to a rectangle of length 2 (seconds) and a height of 1 (meters/second)
2 sec X 1 m/s = 2 meters traveled.

Answer 10

A constant velocity produces a rectangle under the curve. Also, integrating a constant with dt leads simply to "t"... then you evaluate t at both ends of the integral and subtract.

Since velocity is constant, the person also travels 2 meters between seconds 3 and 5, because (using the integral), you evaluate t at t=5 and at t=3 and subtract to get 2. Using the rectangle/geometric method, the length along x is still 2 seconds, and the height is still the constant velocity, 1 m/s, and 2 sec X 1 m/s is again = to 2 meters.