Today, I attempted the third question in the first chapter (Physical Fundamentals of Mechanics) in the book Problems in General Physics by I. E. Irodov. The question goes like this:

A car starts moving rectilinearly, first with acceleration $\omega=5.0\ ms^{-2}$ (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate $\omega$, comes to a stop. The total time of motion equals $\tau=25\ s$. The average velocity during that time is equal to $\langle v\rangle=72\ km\ h^{-1}$. How long does the car move uniformly?

Here's how I solved the problem.

First of all, I drew the velocity-time graph describing the motion of the car, which I provide below:

Velocity-time graph

While accelerating,

Initial velocity $(u)=0$
Acceleration $(a)=\omega$

Assuming the time interval to be $t_0$, we get

Final velocity $(v)=u+at$ $$or,\: v=0+\omega t_0$$ $$or,\: t_0=\frac{v}{\omega}\tag{1}$$

While decelerating,

Initial velocity $(u)=v$
Acceleration $(a)=-\omega$

Assuming the time interval to be $t_2$, we get

Final velocity $(v)=u+at$ $$or,\: 0=v-\omega t_1$$ $$or,\: t_1=\frac{v}{\omega}$$ $$or,\: t_1=t_0\ [from\ (1)]$$

Now, if the car travels with uniform velocity for a time $t_1$, then we have, $$\tau=2t_0+t_1=25\ s\tag{2}$$

We know, average velocity $\langle v\rangle=\frac{total\ displacement}{total\ time}$ $$or,\: \langle v\rangle=\frac{S}{\tau}$$ $$or,\: \frac{72\times 5}{18}=\frac{S}{25}$$ $$or,\: S=500\ m$$

Moreover, $S=Area\ under\ the\ graph$ $$or,\: 500=\frac{1}{2}vt_0+vt_1+\frac{1}{2}vt_2$$ $$or,\: v(t_0+t_1)=500$$ $$or,\: v(\tau-t_0)=500$$ $$or,\: \omega t_0(25-t_0)=500$$ $$or,\: t_0^2-25t_0+100=0$$

Solving it, we get $t_0=5\ s$ as $t_0\neq 20\ s\ (t_0 < \frac{25}{2}\ s)$

Hence, $t_1=25-2\times 5\ s=15\ s$

After putting a lot of effort I find that my perseverance has paid off: I got the correct answer. But thing which I saw on the answer page made me feel that I am an idiot:

$$\Delta t=\tau \sqrt{1-4\langle v \rangle/\omega t}=15\ s$$

It seems as though my approach is unnecessarily lengthy. What I do not understand is:

How did the author arrive at that ridiculously simple expression?

Also, I would like to know if there any more terse or concise approaches to questions like this.


closed as off-topic by ACuriousMind, Martin, John Rennie, Kyle Kanos, yuggib Jun 15 '15 at 9:16

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  • $\begingroup$ It looks like he may have just done exactly the working you did, just without putting in numbers until the end (i.e. his expression looks like the solution to a quadratic, just as in your working) $\endgroup$ – Zephyr Jun 14 '15 at 12:30
  • $\begingroup$ @Zephyr Yes, but how could he have disregarded the invalid value (i.e. the other root) for $\Delta t$? $\endgroup$ – Hungry Blue Dev Jun 14 '15 at 12:35
  • $\begingroup$ or alternatively, notice the average velocity during the acceleration periods is omegat0/2, call the steady velocity u, hence omegat0*t0+vt1=72 (average velocity) and omegat0=v (total velocity) and 2*t0 + t1 =T (total time). I haven't worked this through and about to run out the door but seems like it should work. $\endgroup$ – Zephyr Jun 14 '15 at 12:36
  • 1
    $\begingroup$ how could he have disregarded the invalid value Does "Because he knew the answer before writing it down & is only interested in showing you the right answer without bothering to show the wrong one." count? $\endgroup$ – Kyle Kanos Jun 14 '15 at 15:40

You can easily get that answer by noticing that in when you have constant acceleration the average velocity after a time $t$ is:

\begin{equation} \langle v\rangle_{acc.=w}=\frac{v_{start}+(v_{start}+w t)}{2} \end{equation}

for your case $v_{start}=0$ and then $\langle v\rangle_{acc.=w}=\frac{wt}{2}$. For the deceleration we get the same result.

Now you can write \begin{align} \langle v\rangle\cdot\tau&=\underbrace{\frac{w t}{2}}_{\langle v\rangle_{acceleration}} t + \underbrace{w t}_{v_{uniform}} \underbrace{(\tau -2 t)}_{=:\Delta t} +\underbrace{\frac{w t}{2}}_{\langle v\rangle_{deceleration}} t\\ &=wt\left( t + \Delta t\right) , \end{align} where $\Delta t$ is the time you are looking for. You can easily write $2t=\tau-\Delta t$ in the last equation and then solve: \begin{equation} \langle v\rangle\cdot\tau=\frac{w}{4}(\tau^2-\Delta t^2) \quad\Rightarrow\quad \Delta t=\tau\sqrt{1-\frac{4\langle v\rangle}{w\tau}} \end{equation}


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