# A problem on rectilinear motion in a car [closed]

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:

## 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, yuggibJun 15 '15 at 9:16

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind, Martin, John Rennie, Kyle Kanos, yuggib
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• 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) – Zephyr Jun 14 '15 at 12:30
• @Zephyr Yes, but how could he have disregarded the invalid value (i.e. the other root) for $\Delta t$? – Hungry Blue Dev Jun 14 '15 at 12:35
• 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. – Zephyr Jun 14 '15 at 12:36
• 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? – 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:
$$\langle v\rangle_{acc.=w}=\frac{v_{start}+(v_{start}+w t)}{2}$$
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: $$\langle v\rangle\cdot\tau=\frac{w}{4}(\tau^2-\Delta t^2) \quad\Rightarrow\quad \Delta t=\tau\sqrt{1-\frac{4\langle v\rangle}{w\tau}}$$