# For how long is an objects velocity it's instantaneous velocity at time $t$?

Basically I'm asking if an object's instantaneous velocity at time $t$ is $8m/s$ and its instantaneous velocity at time $t^+$ (idk latex, but basically the t + an infinitely small number) is $10m/s$, then how much has the object traveled in the time between $t$ and $t^+$?

Or in a similar vein, consider an object that travels for $2$ seconds. For $t$ in the set $(0,2]$, its instantaneous velocity is $10m/s$ but at $t = 0$, its instantaneous velocity is $8m/s$. How far does this object travel?

I don't know calculus, but only basic limits. Still, answers based on calculus are still appreciated because I may understand them in a few weeks.

• Your example does not really make sense in practice - physical velocities are never discontinuous. What situation are you imagining when you talk about an object that has a certain velocity at one instant, and then suddenly a not-infinitesimally larger velocity an infinitesimal instant later? – ACuriousMind Aug 19 '17 at 11:04

## 2 Answers

More or less from the definition of the derivative of position with respect to time (in 1D), we have

$$x(t + dt) = x(t) + \frac{dx}{dt}dt = x(t) + v(t)dt$$

In your problem statement, you imply that $t_+ = t + dt$ and so the distance traveled in time $dt$ is $dx = v(t) dt$.

In other words, $v(t_+)$ is irrelevant to $x(t_+)$.

The answer is 20 m.

$s = \int v\cdot dt$ where v=10m/s over the interval (0,2]

$s = 10*t$ (0,2]

s = 20m - 0m = 20m

The instantaneous speed at t=0 of 8m/s is irrelevant because it is not integrated over any amount of time. The continuous function from (0,2]has been defined as v=10m/s and the speed function is not continuous at t=0 and therefore not differentiable at t=0.

The amount of time that the function at t=0 is 8 m/s is undefined and it is meaningless to talk about how long it is at this instantaneous speed. Notice that calculus only works over the interval (0,2] because it is defined by a continuous function v=10 over the specified interval. It could have been defined as some more complex function where the instantaneous speed varied over the interval and calculus would still work and the instantaneous speed would still be meaningful. But in this case the length of time of any instantaneous speed would only have an abstract meaning of an arbitrarily small interval of time surrounding the given time where an average speed could approximate the instantaneous speed.

• Hmm. Ok but what if i modified my example so that instead of 8m/s being at t = 0, it was at t = 0 at exactly 1 instance between 0 and 2. All other instances have a speed of 10m/s. Would that change anything? @mmainville – Typical Highschooler Aug 19 '17 at 3:46
• If by that you mean v=8m/s at some t between 0 and 2 ( let's say t=1) then the function becomes discontinuous and cannot be integrated over the interval (0,2). So we could no longer solve for the distance traveled over the interval (0,2) since the function really would not make any sense around t=1 – mmainville Aug 19 '17 at 3:53
• Please do not provide explicit answers to homework-like problems. – UKH Aug 19 '17 at 3:55