Does area under velocity-time graph represent total distance travelled or total displ

[burgerandcheese]

Hi. I'm new here, and I hope I posted this question in the right section.

Question: Does the area under a velocity-time graph represent the total distance travelled or the total displacement?

Because [FONT=&quot]∫ v dt = s (where v is the velocity function and s is the displacement function).

Here's a question for example:

[/FONT][FONT=&quot]"A particle moves in a straight line and passes through a fixed point A with velocity 2m/s. Its acceleration, a m/s^2, at time t seconds after passing through A is given by a = 15t - 8. [/FONT]

[FONT=&quot]Find the distance travelled by the particle in the fourth second. "

Here's my attempt:

[/FONT][FONT=&quot]v(t) = ∫ (15t - 8)dt = (15/2)t^2 - 8t + c [/FONT]
[FONT=&quot]t = 0, v = 2, c = 2 [/FONT]
[FONT=&quot]v(t) = (15/2)t^2 - 8t + 2

[/FONT]So now if I integrate the velocity function from 3 to 4, I get anxious because if I think of the displacement-time graph, it is a cubic graph and it has turning points.

[FONT=&quot]s(t) = ∫ ((15/2)t^2 - 8t + 2)dt = (15/6)t^3 - 4t^2 + 2t + c [/FONT]
[FONT=&quot]t = 0, s = 0, c = 0 [/FONT]
[FONT=&quot]s(t) = (15/6)t^3 - 4t^2 + 2t [/FONT]

Here's a graph of the displacement function, s(t): https://www.desmos.com/calculator/ytjvhjcksf



Luckily enough, from t = 3 and t = 4 doesn't fall anywhere near the turning points. So I don't have to worry. But what if it does? What if a question asked for the total distance travelled at the tth second, then would it be correct to just simply integrate the velocity function from (t - 1) to t?

Thanks. I hope that made sense..

 

[stapel]

Question: Does the area under a velocity-time graph represent the total distance travelled or the total displacement?

Because ∫ v dt = s (where v is the velocity function and s is the displacement function).

Look at how the variables are defined: "v" is the "velocity" function, and "s" is the "displacement" function. So clearly the integral (being the area under the "velocity" line on the graph) represents the displacement, not the distance travelled.

Think about starting at speed zero, smoothly accelerating to speed V, then smoothly decelerating back to zero, and then in reverse smoothing accelerating to -V, and then decelerating (still in reverse) back to zero (where all the accelerations have the same absolute value A). You'll be back to your original position, so your displacement will have be zero. This makes sense, because the areas of the two halves of the velocity-time graph are equal, other than the second one being the negative of the first one.

The distance travelled, however, will not be zero.

 

[burgerandcheese]

Stapel,

Thank you so much for clarifying. I fully understand it now!