The car cruises for $\Delta t=\tau\sqrt{1-\dfrac{4\langle v\rangle}{a\tau}}=25\sqrt{1-\dfrac{80}{125}}=15\text{ s}$.
(ii) Problem Statement
"A car starts moving rectilinearly, first with acceleration $a=5.0\text{ m/s}^2$ (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate $a$, comes to a stop. The total time of motion equals $\tau=25\text{ s}$. The average velocity during that time is equal to $\langle v\rangle=72\text{ km per hour}$. How long does the car move uniformly?"
Reference Answer Verification (from the supplied solution image):
The reference sets each ramp time to $\dfrac{\tau-\Delta t}{2}$ and writes $\langle v\rangle\tau=2\cdot\tfrac12 a\dfrac{(\tau-\Delta t)^2}{4}+a\dfrac{\tau-\Delta t}{2}\Delta t$. This gives $\Delta t^2=\tau^2-\dfrac{4\langle v\rangle\tau}{a}$ and $\Delta t=15\text{ s}$. Our work matches exactly.
(iii) The Big Idea in Plain Words
The car does three things. It speeds up. It cruises at a steady speed. It slows down to a stop.
Speeding up and slowing down use the same acceleration $a$. So both take the same time and cover the same distance. Call that time $t_r$. Call the cruise time $\Delta t$. Then $2t_r+\Delta t=\tau$, so $t_r=\dfrac{\tau-\Delta t}{2}$.
First change the speed to m/s: $72\text{ km/h}=20\text{ m/s}$. Then total distance is $s=\langle v\rangle\tau=20\times25=500\text{ m}$. Add up the three distances, set them equal to $500\text{ m}$, and solve for $\Delta t$.
(iv) The Solution Viewed in Each Reference Frame
1. Ground or Lab Frame (Stationary Observer)
Origin fixed at the startWhy: both ramps match, so each lasts $t_r=\dfrac{\tau-\Delta t}{2}$.
The top speed is $v_{max}=at_r$. The engine force is $F=ma$ while speeding up and $-ma$ while braking. It is $0$ while cruising.
Why: distance during steady acceleration from rest is $\tfrac12 at_r^2$, and the braking ramp is its mirror image.
$$s=2\cdot\tfrac12 at_r^2+(at_r)\Delta t=\frac{a(\tau-\Delta t)^2}{4}+\frac{a(\tau-\Delta t)\Delta t}{2}$$
Why: simplify, then use $s=\langle v\rangle\tau$.
$$\langle v\rangle\tau=\frac{a(\tau^2-\Delta t^2)}{4}\;\Rightarrow\;\Delta t=\tau\sqrt{1-\frac{4\langle v\rangle}{a\tau}}=25\sqrt{0.36}=15\text{ s}$$
2. Medium Frame (Air or Observer Moving at Steady Speed $u$)
Origin moves at $u$Why: a steady frame shifts every velocity by the same amount and leaves accelerations alone.
Each velocity becomes $v'=v-u$. The acceleration $a$ and all times stay the same.
Why: the car's mean velocity here is the new distance over the same time.
$\langle v\rangle'=\langle v\rangle-u$. The car now starts and ends at speed $-u$, not zero.
Why: to get the answer, return to the true mean $\langle v\rangle=\langle v\rangle'+u$.
The same equation follows, so $\Delta t=15\text{ s}$. Use the sidebar slider $u$ to see the shifted graph.
3. Centre-of-Mass (CM) Frame
This frame does not apply. The car is treated as one point, so it is its own centre of mass. Its velocity and momentum are always $0$ here. The frame accelerates, so a pseudo-force (a "fake" force $-ma$ seen only by an accelerating observer) cancels the real engine force. This shows nothing new, and $\Delta t$ is unchanged.
What Changes and What Stays the Same Across Frames
- What changes: the velocities (by $-u$), the distance covered, and the mean velocity.
- What stays the same: acceleration $a$, every time interval, and the answer $\Delta t=15\text{ s}$.
(v) The Solution by Each Method
Method 1: Graph Method ($v$-$t$ Trapezoid)
Fastest- Why: the $v$-$t$ graph is a trapezoid, and its area is the distance.
The long side has width $\tau$, the short side has width $\Delta t$, and the height is $v_{max}=a\,\dfrac{\tau-\Delta t}{2}$. - Why: area of a trapezoid is height times the average of its two parallel sides.
$\langle v\rangle\tau=v_{max}\,\dfrac{\tau+\Delta t}{2}=\dfrac{a(\tau^2-\Delta t^2)}{4}$. - Why: solve for $\Delta t$.
$\Delta t^2=\tau^2-\dfrac{4\langle v\rangle\tau}{a}=625-400=225$, so $\Delta t=15\text{ s}$.
Method 2: Calculus Method
- Why: acceleration is the rate of change of velocity, so integrate it.
For $0\le t\le t_r$: $v=at$. For $t_r\le t\le t_r+\Delta t$: $v=at_r$. Then $v=at_r-a(t-t_r-\Delta t)$ until $v=0$. - Why: distance is the integral of velocity.
$s=\displaystyle\int_0^{t_r} at\,dt+at_r\Delta t+\int_0^{t_r} (at_r-at)\,dt=at_r^2+at_r\Delta t$. - Why: put in $t_r=\frac{\tau-\Delta t}{2}$ and $s=\langle v\rangle\tau$.
$\langle v\rangle\tau=\dfrac{a(\tau^2-\Delta t^2)}{4}\Rightarrow\Delta t=15\text{ s}$.
Method 3: Force Method
- Why: the free-body diagram shows which forces act along the road.
Weight $mg$ and normal force $N$ cancel. Along the road only the engine force (or the braking force) acts. - Why: Newton's second law turns force into acceleration.
$F=ma$ gives acceleration $+a$ while speeding up, $0$ while cruising, and $-a$ while braking. The mass $m$ cancels. - Why: these accelerations lead to the same distances as before.
Then $s=at_r^2+at_r\Delta t$, which gives $\Delta t=15\text{ s}$.
Method 4: Energy Method
- Why: the work done by the engine equals the gain in kinetic energy (energy of motion).
$ma\,d=\tfrac12 m v_{max}^2$, so each ramp distance is $d=\dfrac{v_{max}^2}{2a}$. - Why: add the two ramps and the cruise.
$s=2d+v_{max}\Delta t=\dfrac{v_{max}^2}{a}+v_{max}\Delta t$ with $v_{max}=a\dfrac{\tau-\Delta t}{2}$. - Why: this is the same equation as before.
$\langle v\rangle\tau=\dfrac{a(\tau^2-\Delta t^2)}{4}$, so $\Delta t=15\text{ s}$.
Method 5: Relative-Motion Method
Not suited. A moving frame only shifts velocities by $u$ and does not shorten the work, so we do not force it.
Fastest: Method 1 (graph). Most reliable in an exam: Method 2 (calculus), since it checks every step. All give $\Delta t=15\text{ s}$.
(vi) Parameter Table
| Symbol | Meaning | Unit | Live value |
|---|---|---|---|
| $a$ | size of the acceleration and braking | m/s² | |
| $\tau$ | total time of motion | s | |
| $\langle v\rangle$ | average velocity over the whole trip | m/s (km/h ÷ 3.6) | |
| $t_r$ | time for each ramp, $(\tau-\Delta t)/2$ | s | |
| $v_{max}$ | top speed, $at_r$ | m/s | |
| $s$ | total distance, $\langle v\rangle\tau$ | m | |
| $\Delta t$ | time of uniform motion (answer) | s |
(vii) Common Exam Traps
- Forgetting to change 72 km/h into 20 m/s. Divide by $3.6$.
- Using $\tau-\Delta t$ as one ramp time. Each ramp is only half of it.
- Writing the trapezoid area with the wrong short side. It is $\Delta t$, not $\tau/2$.
- Taking only the positive root and dropping the check. $\Delta t$ must be between $0$ and $\tau$.
- Forgetting the limit $\langle v\rangle\le\dfrac{a\tau}{4}$. Above it there is no solution.
Verified: the live numbers, all frames and all methods give $\Delta t=15\text{ s}$ with the default values, with units in m, s and m/s.