Find time from distance-speed graph

[Metaphysics]

In audio application i can automate playrate envelope (playhead speed)
1 = original speed, 2 = 2 x original speed, 3 = 3 x original speed etc..

Now i need to calculate time it takes for playhead to travel each section

its easy to calculate time for section 1, where speed is constant
time = section 1 len/speed = 1/1 = 1 sec

but i don't know how to find time for section 2, section 3

can you help?

 

[Doc Al]

but i don't know how to find time for section 2, section 3

I don't really know what you're calculating. Nonetheless, it looks like the speed changes linearly in those sections, so you should be able to use the average speed.

 

[Doc Al]

I think my previous response was a bit glib, seeing as the horizontal axis is distance and not time. But for each section, you can write the speed as a linear function of distance, then rearrange and integrate.

 

[kuruman]

I would follow a different approach from @Doc Al. Start with the standard transformation for the acceleration.
$$a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$$In sections 2 and 3 ##\dfrac{dv}{dx}=\text{const.}=\alpha##. You can easily get the slope ##\alpha## in each section from the graph. Then $$\frac{dv}{dt}=\alpha ~v.$$Can you solve this simple differential equation to find ##v(t)##? If so, then do it, but don't forget the initial condition at the beginning of each section. Then invert the equation to find ##t## as a function of ##v## and you're done. If you cannot solve the differential equation, we are here to help.

 

[Doc Al]

Good stuff @kuruman.

I'll outline the approach I mentioned above (which was just the first that occurred to me). Start by expressing the velocity as a function of distance, which should be trivial since it's a straight line. For example, using the standard form for a straight line:
$$ v(x) = mx + b$$
That gives you the following simple differential equation, which you can solve:
$$v = \frac{dx}{dt} = mx + b$$

Either way, it's all good. (For fun, solve it both ways and compare.)