Chapter 2 · Motion in a Straight Line

2.2 Acceleration and Motion Graphs

Acceleration is the rate of change of velocity, $a=\dfrac{dv}{dt}$. The slopes and areas of x–t, v–t and a–t graphs tie position, velocity and acceleration together.

Stand in a metro train with your eyes closed. While it cruises at a steady $80\ \text{km/h}$ you feel nothing at all — a cup of water on the floor would not spill. But in the few seconds when the train pulls away or brakes into a station, you lurch and reach for the handrail. Your body does not sense position or velocity; it senses changes of velocity. That change per second is the acceleration, and it is what engineers limit for comfort (around $1\ \text{m/s}^2$ in a metro) and what brakes, engines and — in Chapter 4 — forces produce.

In this section we define acceleration as a rate, $a=\dfrac{dv}{dt}$, exactly as velocity was defined from position. We learn to tell speeding up from slowing down (it is not just the sign of $a$), read and sketch the three motion graphs — x–t, v–t and a–t — and discover the second big idea of calculus: the area under a velocity–time graph is the displacement.

Acceleration

📘 Definition 2.5 — Average and instantaneous acceleration

If the velocity changes from $v_1$ at time $t_1$ to $v_2$ at time $t_2$, the average acceleration over the interval is

$$\bar a=\frac{\Delta v}{\Delta t}=\frac{v_2-v_1}{t_2-t_1}.$$

The instantaneous acceleration (simply "acceleration") is the limit as the interval shrinks,

$$a=\lim_{\Delta t\to0}\frac{\Delta v}{\Delta t}=\frac{dv}{dt}=\frac{d^2x}{dt^2},$$

the slope of the tangent to the v–t graph. Its SI unit is $\text{m/s}^2$ (metres per second, per second), dimensions $[\text{L}\,\text{T}^{-2}]$. In one dimension its sign gives its direction along the axis. When $a$ is the same at every instant the motion is uniformly accelerated.

The notation $\dfrac{d^2x}{dt^2}$ ("second derivative") just means: differentiate $x$ once to get $v$, then differentiate $v$ to get $a$. With Result 2.1 this is quick for any polynomial motion.

✏️ Example 1 — From zero to $100\ \text{km/h}$
A car reaches $100\ \text{km/h}$ from rest in $8.0\ \text{s}$. A second car brakes from $72\ \text{km/h}$ to rest in $4.0\ \text{s}$. Find each average acceleration, taking the direction of motion as positive.
Convert $100\ \text{km/h}=\dfrac{100}{3.6}=27.8\ \text{m/s}$; $\ 72\ \text{km/h}=20\ \text{m/s}$.
First car $\bar a=\dfrac{27.8-0}{8.0}=3.5\ \text{m/s}^2$.
Second car $\bar a=\dfrac{0-20}{4.0}=-5.0\ \text{m/s}^2$: the acceleration points opposite to the motion. A negative acceleration while moving forward is often called a deceleration or retardation of $5.0\ \text{m/s}^2$.
Check Units: $(\text{m/s})/\text{s}=\text{m/s}^2$ ✓. Every second, the first car gains $3.5\ \text{m/s}$ ($12.5\ \text{km/h}$) of speed.

Does a negative acceleration mean slowing down? Not always. Velocity and acceleration each carry a sign, and what matters is whether they agree.

📐 Result 2.2 — Slopes of motion graphs; speeding up or slowing down

(i) The velocity is the slope of the x–t graph, $v=\dfrac{dx}{dt}$; the acceleration is the slope of the v–t graph, $a=\dfrac{dv}{dt}$.

(ii) At any instant when $v\ne0$: the body is speeding up if $v$ and $a$ have the same sign, and slowing down if they have opposite signs. (If $a=0$ the speed is momentarily not changing.)

(iii) On the x–t graph, $a\gt0$ shows as a curve bending upwards (concave up) and $a\lt0$ as a curve bending downwards.

✏️ Example 2 — When is it speeding up?
A particle moves with $x=2t^3-9t^2+12t+1$ ($x$ in m, $t$ in s). (a) Find $v(t)$ and $a(t)$. (b) Find the intervals during which it is speeding up and those in which it is slowing down, for $t\ge0$.
(a) $v=\dfrac{dx}{dt}=6t^2-18t+12=6(t-1)(t-2)\ \text{m/s}$; $\ a=\dfrac{dv}{dt}=12t-18\ \text{m/s}^2$.
Signs $v\gt0$ for $t\lt1$, $v\lt0$ for $1\lt t\lt2$, $v\gt0$ for $t\gt2$. $\ a\lt0$ for $t\lt1.5$, $a\gt0$ for $t\gt1.5$.
(b) Same signs (speeding up): $1\lt t\lt1.5\ \text{s}$ (both negative) and $t\gt2\ \text{s}$ (both positive). Opposite signs (slowing down): $0\le t\lt1\ \text{s}$ and $1.5\lt t\lt2\ \text{s}$.
Check At $t=0$: $v=12\ \text{m/s}$, $a=-18\ \text{m/s}^2$ — moving forward but braking, so the speed falls to $0$ at $t=1\ \text{s}$ ✓. On $1\lt t\lt1.5$ the acceleration is negative and the particle speeds up — a negative acceleration that is not a slowing down.

Motion graphs: x–t, v–t and a–t

The three graphs describe one motion in three ways. Each graph is the slope of the one above it, so their shapes are linked: a straight x–t graph goes with a horizontal v–t graph and a zero a–t graph; a parabolic x–t graph goes with a straight v–t graph and a horizontal a–t graph.

Motionx–t graphv–t grapha–t graph
at resthorizontal lineon the $t$-axis ($v=0$)on the $t$-axis
uniform velocity $v$straight line, slope $v$horizontal line at $v$on the $t$-axis
constant $a\gt0$, starting from restparabola bending upstraight line through origin, slope $a$horizontal line at $a$
braking to rest ($v\gt0$, $a\lt0$)curve flattening to horizontalstraight line falling to the $t$-axishorizontal line below the axis
thrown up, comes back downparabola with a peakstraight line crossing the $t$-axishorizontal line at $-g$
Interactive 1 — A Cart with Live x–t, v–t and a–t Graphs

The cart starts at $x=0$ with initial velocity $u$ and constant acceleration $a$; its three graphs are drawn as it moves (in real time, $8\ \text{s}$). The green and red shading under the v–t graph is its signed area — compare it with $x$. Try $u\gt0$ with $a\lt0$: the cart slows, stops and comes back.

8 m/s−2 m/s²
time $t$—
position $x$—
velocity $v$—
acceleration $a$—
the cart is…—
✏️ Example 3 — Matching the graphs of a lift
A lift starts from rest at the ground floor, speeds up uniformly to $2.0\ \text{m/s}$ in $2.0\ \text{s}$, moves steadily for $6.0\ \text{s}$, then slows uniformly to rest in $2.5\ \text{s}$. Taking upwards as positive, describe the v–t, a–t and x–t graphs, giving the acceleration in each stage.
v–t Straight segments joining $(0,0)$, $(2.0\ \text{s},\,2.0\ \text{m/s})$, $(8.0\ \text{s},\,2.0\ \text{m/s})$, $(10.5\ \text{s},\,0)$ — a trapezium.
a–t Slopes: $\dfrac{2.0}{2.0}=+1.0\ \text{m/s}^2$ for $0$–$2\ \text{s}$; $0$ for $2$–$8\ \text{s}$; $\dfrac{0-2.0}{2.5}=-0.80\ \text{m/s}^2$ for $8$–$10.5\ \text{s}$. Three horizontal steps.
x–t A parabola bending up ($a\gt0$), joining smoothly onto a straight line of slope $2.0\ \text{m/s}$, then a parabola bending down that flattens to horizontal at $10.5\ \text{s}$ (the lift is at rest at the top).
Note The x–t graph has no corners: the velocity (its slope) never jumps. The a–t graph does jump — real lifts smooth these jumps to avoid a jolt.

Area under the v–t graph = displacement

If a body moves with constant velocity $v$ for a time $\Delta t$, its displacement is $v\,\Delta t$ — the area of the rectangle under its horizontal v–t graph. When $v$ changes, cut the time into many thin strips of width $\Delta t$. In each strip the velocity is nearly constant, so the displacement in it is nearly the area of a thin rectangle. Adding the strips and making them thinner and thinner, the sum becomes exactly the area under the v–t curve. Strips below the $t$-axis ($v\lt0$) give negative displacements.

📐 Result 2.3 — Areas under motion graphs

(i) The displacement between $t_1$ and $t_2$ equals the signed area between the v–t graph and the $t$-axis, written as an integral:

$$x_2-x_1=\int_{t_1}^{t_2}v\,dt .$$

Area above the axis counts positive, area below counts negative. The path length is the total area with every part counted positive.

(ii) In the same way, the change in velocity equals the signed area under the a–t graph:

$$v_2-v_1=\int_{t_1}^{t_2}a\,dt .$$

(iii) Since $\dfrac{d}{dt}\!\left(\dfrac{t^{n+1}}{n+1}\right)=t^n$, the area under $t^n$ from $0$ to $T$ is $\dfrac{T^{n+1}}{n+1}$ for $n\ge0$ (you will prove this in maths). For straight-line graphs, simply use triangles, rectangles and trapeziums.

✏️ Example 4 — How high does the lift go?
For the lift of Example 3, find (a) the height it rises, (b) its average velocity for the whole trip, (c) the height it has risen after $5.0\ \text{s}$.
(a) Area $=\tfrac12(2.0)(2.0)+(6.0)(2.0)+\tfrac12(2.5)(2.0)=2.0+12+2.5=16.5\ \text{m}$.
(b) $\bar v=\dfrac{16.5\ \text{m}}{10.5\ \text{s}}=1.6\ \text{m/s}$ upwards.
(c) Area up to $5.0\ \text{s}$: $2.0+(3.0)(2.0)=8.0\ \text{m}$.
Check Trapezium formula for the whole graph: $\tfrac12(\text{sum of parallel sides})\times\text{height}=\tfrac12(10.5+6.0)(2.0)=16.5\ \text{m}$ ✓.
✏️ Example 5 — A toy car that reverses
A remote-control car has $v=4.0-2.0\,t$ ($v$ in m/s, $t$ in s) for $0\le t\le5.0\ \text{s}$. Using areas under the v–t graph, find its displacement and the path length in these $5.0\ \text{s}$. Check with $x(t)$.
Graph A straight line from $(0,\,4.0)$ to $(5.0,\,-6.0)$, crossing the axis at $t=2.0\ \text{s}$ — the car turns back there.
Areas Above: $\tfrac12(2.0)(4.0)=+4.0\ \text{m}$. Below: $\tfrac12(3.0)(-6.0)=-9.0\ \text{m}$.
Answer Displacement $=4.0-9.0=-5.0\ \text{m}$; path length $=4.0+9.0=13\ \text{m}$.
Check $x=4.0t-t^2$ has $\dfrac{dx}{dt}=4.0-2.0t$ ✓ and $x(5.0)=20-25=-5.0\ \text{m}$ ✓; $x(2.0)=4.0\ \text{m}$, so the path is $4.0+9.0=13\ \text{m}$ ✓.
✏️ Example 6 — Velocity from an a–t graph
A rocket sled starts from rest. Its acceleration is $+3.0\ \text{m/s}^2$ for $4.0\ \text{s}$, then $-2.0\ \text{m/s}^2$ for the next $6.0\ \text{s}$. Find its velocity at $t=4.0\ \text{s}$ and $t=10.0\ \text{s}$, and the total distance covered.
a–t area $v(4.0)=0+(3.0)(4.0)=12\ \text{m/s}$; $\ v(10.0)=12+(-2.0)(6.0)=0$.
v–t So the v–t graph is a triangle: up from $(0,0)$ to $(4.0,\,12)$, down to $(10.0,\,0)$; $v$ never becomes negative.
Distance Area $=\tfrac12(10.0)(12)=60\ \text{m}$.
Check Average velocity $6.0\ \text{m/s}$, half the peak — as it must be for a triangle.
⚠️ Common mistakes with motion graphs

1. Slope or height? On a v–t graph the height is the velocity, the slope is the acceleration and the area is the displacement. On an x–t graph the height is the position and the slope is the velocity; its "area" means nothing.

2. Zero velocity does not mean zero acceleration. At the turning point of Example 5 ($t=2.0\ \text{s}$) the car is momentarily at rest but its acceleration is still $-2.0\ \text{m/s}^2$ — otherwise it would never start moving back.

3. Negative acceleration is not always "slowing down". Compare the signs of $v$ and $a$ (Result 2.2).

4. Distance needs the turning points. Split the v–t graph where it crosses the $t$-axis and add the areas without signs.

Interactive 2 — Area under v–t Builds the x–t Graph

As time runs, the area under the v–t graph (top) is shaded — green above the axis, red below. The lower graph plots the net shaded area against time: it is exactly the x–t graph. Watch it reach a peak at the instant $v$ crosses zero. Press Play or drag the time slider.

time $t$—
velocity $v$—
area above axis—
area below axis—
displacement / path length—

Exercises

I
Acceleration. Positions in metres, times in seconds.
  • 1 A bus speeds up from $10\ \text{m/s}$ to $25\ \text{m/s}$ in $6.0\ \text{s}$. Find its average acceleration.
  • 2 A scooter brakes from $54\ \text{km/h}$ to rest in $3.0\ \text{s}$. Find its average acceleration (direction of motion positive).
  • 3 A ball hits a wall at $12\ \text{m/s}$ and rebounds along the same line at $8.0\ \text{m/s}$; the contact lasts $0.040\ \text{s}$. Taking the direction towards the wall as positive, find the average acceleration during contact.
  • 4 $v=3.0\,t^2-2.0\,t$ (m/s). Find the acceleration at $t=2.0\ \text{s}$.
  • 5 $x=5.0+4.0\,t-2.0\,t^2$. Find $v(t)$ and $a(t)$. Is the acceleration uniform?
  • 6 For the body of part 5, when is it speeding up and when is it slowing down ($t\ge0$)?
  • 7 True or false, with a reason: (a) a body that is momentarily at rest must have zero acceleration; (b) a body with negative acceleration is always slowing down.
1 $\bar a=\dfrac{25-10}{6.0}=2.5\ \text{m/s}^2$.
2 $54\ \text{km/h}=15\ \text{m/s}$: $\bar a=\dfrac{0-15}{3.0}=-5.0\ \text{m/s}^2$ (a retardation of $5.0\ \text{m/s}^2$).
3 $\Delta v=(-8.0)-(+12)=-20\ \text{m/s}$; $\bar a=\dfrac{-20}{0.040}=-500\ \text{m/s}^2$, i.e. $500\ \text{m/s}^2$ directed away from the wall. (Using $12-8.0$ is the classic error.)
4 $a=\dfrac{dv}{dt}=6.0\,t-2.0=10\ \text{m/s}^2$.
5 $v=4.0-4.0\,t\ \text{m/s}$, $a=-4.0\ \text{m/s}^2$ at all times: yes, uniform.
6 $v\gt0$ for $t\lt1.0\ \text{s}$, $v\lt0$ after; $a\lt0$ always. Slowing down for $0\le t\lt1.0\ \text{s}$, speeding up for $t\gt1.0\ \text{s}$ (moving in the $-x$ direction).
7 (a) False: at a turning point $v=0$ but $a\ne0$ (a ball at the top of its flight has $a=9.8\ \text{m/s}^2$ downwards). (b) False: if $v$ is also negative the body speeds up.
II
Motion graphs. Describe or sketch the graphs asked for.
  • 1 Sketch the x–t, v–t and a–t graphs of a body at rest at $x=+5\ \text{m}$.
  • 2 A body starts at $x=+12\ \text{m}$ and moves with a constant velocity of $-3.0\ \text{m/s}$. Sketch its three graphs. When does it pass the origin?
  • 3 A car starts from rest with $a=2.0\ \text{m/s}^2$ for $5.0\ \text{s}$. Sketch its v–t graph and find its velocity and position (from $x=0$) at $5.0\ \text{s}$.
  • 4 A v–t graph is a straight line from $(0,\,12\ \text{m/s})$ to $(6.0\ \text{s},\,0)$. Find the acceleration and describe the x–t graph.
  • 5 Which feature of a v–t graph shows that a body turns back? Which feature of an x–t graph?
  • 6 An x–t graph is a parabola opening downwards with its vertex at $t=3\ \text{s}$. Give the signs of $v$ and $a$ before and after $t=3\ \text{s}$. Is the body speeding up or slowing down in each part?
  • 7 A v–t graph is a horizontal line at $v=-4.0\ \text{m/s}$. Describe the x–t and a–t graphs.
1 x–t: horizontal line at $5\ \text{m}$; v–t and a–t: along the $t$-axis.
2 x–t: straight line from $12\ \text{m}$ with slope $-3.0\ \text{m/s}$; v–t: horizontal at $-3.0\ \text{m/s}$; a–t: zero. It passes the origin when $12-3.0t=0$, at $t=4.0\ \text{s}$.
3 v–t: straight line from the origin, slope $2.0$. $v=2.0\times5.0=10\ \text{m/s}$; $x$ = triangle area $=\tfrac12(5.0)(10)=25\ \text{m}$.
4 $a=\dfrac{0-12}{6.0}=-2.0\ \text{m/s}^2$. The x–t graph rises ever less steeply (concave down) and becomes horizontal at $t=6.0\ \text{s}$, at $x_0+36\ \text{m}$.
5 v–t: the graph crosses the $t$-axis (the sign of $v$ changes). x–t: a peak or valley (horizontal tangent, then the slope changes sign).
6 Concave down: $a\lt0$ throughout. Before $3\ \text{s}$, $v\gt0$: slowing down. After, $v\lt0$: speeding up (in the $-x$ direction).
7 x–t: straight line of slope $-4.0\ \text{m/s}$ (falling); a–t: zero.
III
Areas under graphs.
  • 1 A v–t graph rises uniformly from $0$ to $15\ \text{m/s}$ in $5.0\ \text{s}$, stays constant for $10\ \text{s}$, then falls uniformly to zero in $3.0\ \text{s}$. Find the total displacement and the average velocity.
  • 2 A metro train accelerates from rest at $1.2\ \text{m/s}^2$ for $15\ \text{s}$, cruises, then brakes at $1.5\ \text{m/s}^2$ to stop. The stations are $1.20\ \text{km}$ apart. Find the braking time, the cruising time and the total time.
  • 3 $v=6.0-2.0\,t$ (m/s) for $0\le t\le5.0\ \text{s}$. Find the displacement and the path length.
  • 4 A body starts from rest. Its a–t graph is $+2.0\ \text{m/s}^2$ for $0$–$3.0\ \text{s}$, zero for $3.0$–$5.0\ \text{s}$ and $-1.5\ \text{m/s}^2$ for $5.0$–$9.0\ \text{s}$. Find its velocity at $3.0$, $5.0$ and $9.0\ \text{s}$.
  • 5 For the body of part 4, find the total displacement.
  • 6 A body has initial velocity $-5.0\ \text{m/s}$; the area under its a–t graph over the next $4.0\ \text{s}$ is $+12\ \text{m/s}$. Find its final velocity.
  • 7 A data logger records $v=0,\ 3.0,\ 5.0,\ 6.0,\ 6.5\ \text{m/s}$ at $t=0,\ 2,\ 4,\ 6,\ 8\ \text{s}$. Estimate the displacement in $8\ \text{s}$ using trapeziums.
  • 8 Explain how the area under a v–t graph can be negative, and how you then find the distance travelled.
1 $\tfrac12(5.0)(15)+(10)(15)+\tfrac12(3.0)(15)=37.5+150+22.5=210\ \text{m}$; $\bar v=\dfrac{210}{18}=11.7\approx12\ \text{m/s}$.
2 Peak $v=1.2\times15=18\ \text{m/s}$. Braking time $\dfrac{18}{1.5}=12\ \text{s}$. Areas: speeding up $\tfrac12(15)(18)=135\ \text{m}$, braking $\tfrac12(12)(18)=108\ \text{m}$; cruising $1200-243=957\ \text{m}$ takes $\dfrac{957}{18}=53\ \text{s}$. Total $15+53+12=80\ \text{s}$.
3 Crosses zero at $t=3.0\ \text{s}$; $v(5.0)=-4.0\ \text{m/s}$. Areas $+\tfrac12(3.0)(6.0)=+9.0\ \text{m}$ and $-\tfrac12(2.0)(4.0)=-4.0\ \text{m}$: displacement $+5.0\ \text{m}$, path length $13\ \text{m}$.
4 $v(3.0)=6.0\ \text{m/s}$; $v(5.0)=6.0\ \text{m/s}$; $v(9.0)=6.0-1.5\times4.0=0$.
5 v–t areas: $\tfrac12(3.0)(6.0)+(2.0)(6.0)+\tfrac12(4.0)(6.0)=9.0+12+12=33\ \text{m}$.
6 $v=-5.0+12=+7.0\ \text{m/s}$.
7 $2\left[\tfrac{0+3.0}{2}+\tfrac{3.0+5.0}{2}+\tfrac{5.0+6.0}{2}+\tfrac{6.0+6.5}{2}\right]=3.0+8.0+11+12.5=34.5\approx35\ \text{m}$.
8 Where $v\lt0$ the graph is below the axis and the displacement in that time is negative. For distance, find where $v=0$, compute each area separately and add their magnitudes.
IV
Calculus and modelling.
  • 1 A body starts from $x=0$ with $v=3.0\,t^2$ (m/s). Using Result 2.3(iii), find its displacement in the first $2.0\ \text{s}$.
  • 2 $x=0.20\,t^3$. Find $v$ and $a$ at $t=5.0\ \text{s}$. Is the acceleration uniform?
  • 3 A sprinter speeds up uniformly from rest to $11\ \text{m/s}$ in $4.0\ \text{s}$ and then holds that speed. Find her time for $100\ \text{m}$.
  • 4 A lift must travel $40\ \text{m}$ from rest to rest. Its top speed is $2.5\ \text{m/s}$ and it speeds up and slows down at $1.0\ \text{m/s}^2$. Find the shortest trip time.
  • 5 From rest, a body has $a=2.0\,t$ (m/s²) for $0\le t\le3.0\ \text{s}$. Use the a–t graph to find its velocity at $3.0\ \text{s}$.
  • 6 A body with $x(0)=0$ has $v=8.0-2.0\,t$. Write $x(t)$ (check it by differentiating) and find $x$ at $t=6.0\ \text{s}$. Confirm with areas.
1 Area under $3.0\,t^2$ from $0$ to $2.0$: $3.0\times\dfrac{2.0^3}{3}=8.0\ \text{m}$.
2 $v=0.60\,t^2=15\ \text{m/s}$; $a=1.2\,t=6.0\ \text{m/s}^2$. Not uniform — $a$ grows with $t$.
3 First $4.0\ \text{s}$: area $\tfrac12(4.0)(11)=22\ \text{m}$. Remaining $78\ \text{m}$ at $11\ \text{m/s}$: $7.09\ \text{s}$. Total $11.1\ \text{s}$.
4 Speeding up takes $2.5\ \text{s}$ and covers $\tfrac12(2.5)(2.5)=3.125\ \text{m}$; the same for slowing down. Cruise $40-6.25=33.75\ \text{m}$ at $2.5\ \text{m/s}$: $13.5\ \text{s}$. Total $18.5\ \text{s}$.
5 The a–t graph is a straight line from $0$ to $6.0\ \text{m/s}^2$: triangle area $\tfrac12(3.0)(6.0)=9.0\ \text{m/s}$, so $v=9.0\ \text{m/s}$.
6 $x=8.0t-t^2$ (then $dx/dt=8.0-2.0t$ ✓); $x(6.0)=48-36=12\ \text{m}$. Areas: $+\tfrac12(4.0)(8.0)=16$ and $-\tfrac12(2.0)(4.0)=-4.0$; net $12\ \text{m}$ ✓.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks A lift starts from rest, speeds up uniformly to $3.0\ \text{m/s}$ in $2.0\ \text{s}$, moves steadily for $8.0\ \text{s}$, then slows uniformly to rest in $3.0\ \text{s}$. Draw its v–t graph, find its acceleration in each stage, and find the distance it travels.

Graph Trapezium through $(0,0)$, $(2.0,\,3.0)$, $(10.0,\,3.0)$, $(13.0,\,0)$. [1]
Accelerations $+1.5\ \text{m/s}^2$, $0$, $-1.0\ \text{m/s}^2$ (slopes). [1]
Distance $\tfrac12(2.0)(3.0)+(8.0)(3.0)+\tfrac12(3.0)(3.0)=3.0+24+4.5=31.5\ \text{m}$. [1]
Exam tip Label axes with quantities and units, and mark the corner values on the graph; examiners give credit for them.

E2. NEET The v–t graph of a particle is a straight line with negative slope that crosses the time axis. Its a–t graph is (a) a horizontal line below the time axis (b) a horizontal line above the time axis (c) a straight line with negative slope (d) the time axis itself

Idea $a$ is the slope of the v–t graph: constant and negative.
Answer (a).
Exam tip (c) copies the v–t shape. Crossing the axis changes the sign of $v$, not of $a$ — the particle turns back with the same acceleration.

E3. JEE Main A particle moves with $x=t^3-6t^2+9t$ ($x$ in m, $t$ in s). Its acceleration at the instants when it is momentarily at rest is (a) zero at both instants (b) $-6\ \text{m/s}^2$ and $+6\ \text{m/s}^2$ (c) $+6\ \text{m/s}^2$ at both instants (d) $-12\ \text{m/s}^2$ and $+12\ \text{m/s}^2$

Rest $v=3t^2-12t+9=3(t-1)(t-3)=0$ at $t=1\ \text{s}$ and $t=3\ \text{s}$.
Acceleration $a=6t-12$: $a(1)=-6\ \text{m/s}^2$, $a(3)=+6\ \text{m/s}^2$.
Answer (b).
Exam tip (a) assumes "at rest means no acceleration" — the trap. At a turning point the acceleration points back towards where the body came from.

E4. NEET A particle starts from rest. Its acceleration is $4\ \text{m/s}^2$ for $2\ \text{s}$ and then $-2\ \text{m/s}^2$ for the next $4\ \text{s}$. Its velocity at $t=6\ \text{s}$ is (a) $0$ (b) $8\ \text{m/s}$ (c) $16\ \text{m/s}$ (d) $-8\ \text{m/s}$

Areas Area under a–t: $(4)(2)+(-2)(4)=8-8=0$.
Answer (a): $v=0+0=0$.
Exam tip (b) is the velocity at $2\ \text{s}$; (c) adds the areas without signs.

E5. JEE Advanced (One or more correct) A particle starting at $x=0$ has $v=6-3t$ ($v$ in m/s, $t$ in s) for $0\le t\le4\ \text{s}$. Which statements are correct? (A) Its displacement over the $4\ \text{s}$ is zero. (B) The distance it covers is $12\ \text{m}$. (C) It is speeding up for $2\lt t\lt4\ \text{s}$. (D) Its acceleration is zero at $t=2\ \text{s}$.

Graph Straight line from $(0,\,6)$ to $(4,\,-6)$, crossing zero at $t=2\ \text{s}$; $a=-3\ \text{m/s}^2$ throughout.
(A), (B) Areas $+\tfrac12(2)(6)=+6\ \text{m}$ and $-6\ \text{m}$: displacement $0$ ✓, distance $12\ \text{m}$ ✓.
(C), (D) For $2\lt t\lt4$, $v\lt0$ and $a\lt0$: speeding up ✓. At $t=2\ \text{s}$, $v=0$ but $a=-3\ \text{m/s}^2$ ✗.
Answer (A), (B), (C).

E6. Board · Case study A metro train leaves a station from rest, accelerates uniformly to $20\ \text{m/s}$ in $20\ \text{s}$, runs at this speed for $40\ \text{s}$, and then decelerates uniformly to rest at the next station in $16\ \text{s}$. (i) Find the distance between the stations. [2] (ii) Find the deceleration. [1] (iii) Find the average speed between the stations. [1]

(i) Area of the v–t trapezium: $\tfrac12(20)(20)+(40)(20)+\tfrac12(16)(20)=200+800+160=1160\ \text{m}$. [2]
(ii) $\dfrac{20}{16}=1.25\ \text{m/s}^2\approx1.2\ \text{m/s}^2$ (retardation). [1]
(iii) $\dfrac{1160\ \text{m}}{76\ \text{s}}=15\ \text{m/s}$. [1]
Exam tip In (ii), $1.25$ rounds to $1.2$ by the round-half-to-even rule; writing $1.25\ \text{m/s}^2$ is also accepted, since the data are exact here.
🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. Can a phone's accelerometer alone tell you how high a lift has travelled?

Why it matters. Lift and railway engineers design ride profiles so that acceleration stays comfortable (typically around $1\ \text{m/s}^2$) and changes gradually. Inertial navigation — used in aircraft, submarines, drones and phones when GPS is lost — works on exactly the idea of this section: integrate acceleration to get velocity, and velocity to get position. Its weakness is drift: a small constant error in $a$ grows into a large error in position.

What to do.

  1. Install a free sensor app (for example phyphox) and record the vertical acceleration (the "linear acceleration" sensor, gravity removed) during a lift ride of several floors, holding the phone flat and still.
  2. Sketch the a–t graph. Estimate the areas of the positive and negative pulses; they should be nearly equal and opposite. Why?
  3. From the first area find the top speed of the lift; from the v–t graph you construct, find the height travelled. Compare with the number of floors × the height of one floor.
  4. Suppose the sensor has a constant error of $0.02\ \text{m/s}^2$. Using areas, find the false velocity and the false displacement it produces after $10\ \text{s}$ and after $60\ \text{s}$.

Tools: Smartphone with a sensor app, a building with a lift, graph paper or a spreadsheet.

Hand in: Your a–t and v–t sketches, the estimated height with an uncertainty, and a paragraph on why navigation systems cannot rely on accelerometers alone.

Going further. Look up "jerk" (the rate of change of acceleration) and why lift and roller-coaster designers limit it as well as the acceleration itself.

Instructor note: For step 4, a constant error $\delta a$ gives $\delta v=\delta a\,t$ and $\delta x=\tfrac12\delta a\,t^2$: $0.2\ \text{m/s}$ and $1\ \text{m}$ after $10\ \text{s}$; $1.2\ \text{m/s}$ and $36\ \text{m}$ after $60\ \text{s}$ — the quadratic growth is the key insight. Typical lift data: pulses of about $\pm0.8$–$1.2\ \text{m/s}^2$ lasting $1.5$–$2.5\ \text{s}$; students' heights usually agree with the building to within $10$–$20\%$.

🧪 Self-check quiz

0/5 correct

Q1. The acceleration of a body is given by

Q2. At some instant a body has $v=-5\ \text{m/s}$ and $a=-2\ \text{m/s}^2$. It is

Q3. A body starts from $x=0$ with $v=2t$ (SI units). Its position at $t=4\ \text{s}$ is

Q4. At the instant a body is momentarily at rest and turning back, its acceleration

Q5. A body moving at $-2\ \text{m/s}$ has a constant acceleration of $+3\ \text{m/s}^2$ for $4\ \text{s}$. Its final velocity is

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