This may be a stupid question, but in an assignment I was given recently I was given an equation for acceleration and told to use integration to work out the velocity after 8 seconds assuming that the device started from rest.

The equation for acceleration looked a little like this.

$$a = \frac{5t}{4} - 1$$

Is there a advantage to using integration over substituting into the formula $v=u+at$?

Like so.

\begin{align} v & = u + (\frac{5t}{4}-1)t \\\\ & = u + \frac{5t^2}{4} - t \end{align}

  • $\begingroup$ $v=u+at$ assumes acceleration is constant. Here it is not, it is linear. You can get away in linear cases by using the average acceleration instead though, I.e. use $a=4$ in the equation. $\endgroup$ – Macavity May 21 '17 at 7:55
  • $\begingroup$ @Macavity Thank you. The answer below explained it to me as well. Also it's helpful to know the average can be used if it is linear $\endgroup$ – Dan May 21 '17 at 7:56
  • $\begingroup$ $a(0)=-1, a(8)=9$ using the linear relation given. Take the average $(-1+9)/2=4$. $\endgroup$ – Macavity May 21 '17 at 10:36

Yup, the main advantage is that integration gives you the right answer, while substituting in doesn't! More specifically, the formula $v = u + at$ applies in a constant acceleration problem, when $a$ doesn't depend on $t$. But here, $a$ does depend on $t$, so the formula doesn't apply.

The integration solution would look like: $$\begin{align} \frac{dv}{dt} &= \frac{5t}{4} - 1 \\ \int_0^8\frac{dv}{dt} \ dt &= \int_0^8\frac{5t}{4} - 1 \ dt\\ v(8) - v(0) &= \left[\frac{5t^2}{8} - t \right]_0^8\\ v(8) &= \frac{5\cdot 64}{8} - 8 \\ v(8) &= 32 \\ \end{align}$$

  • $\begingroup$ Ah. Thank you very much. I didn't know that. I'll accept your answer when it lets me $\endgroup$ – Dan May 21 '17 at 7:48
  • 1
    $\begingroup$ @Dan Yes, that's exactly what it is, the derivative with respect to time. $\endgroup$ – B. Mehta May 21 '17 at 7:50

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