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Physics · Motion in a Straight Line

Acceleration

Acceleration tells you how fast velocity is changing. The signs trip students up — a particle with negative acceleration can be speeding up, a particle at rest can be accelerating. This deep-dive nails the definitions, spells out the four sign-cases of $v$ and $a$, reads the v–t graph slope-by-slope, and walks through NCERT-style worked examples. NEET 2018 and 2016 PYQs sit at the foot of the page.

Why we need a separate quantity called acceleration

The subtopics on position and velocity describe where a particle is and how fast that location is changing. Real motion is rarely uniform — a car pulls away from a light, a stone falls faster the longer it falls. To describe how velocity itself changes with time, we need a third quantity: acceleration.

NCERT §2.3 traces this back to Galileo. His free-fall experiments showed that for a falling body, the rate of change of velocity with time is constant, while the rate of change with distance is not. Acceleration is therefore defined per unit time, not per unit distance. SI unit: $\text{m s}^{-2}$.

"Galileo concluded that the rate of change of velocity with time is a constant of motion for all objects in free fall. … This led to the concept of acceleration as the rate of change of velocity with time."

NCERT Class 11 Physics, §2.3

Average acceleration

The average acceleration $\bar{a}$ over a time interval is the change in velocity divided by the time interval. If $v_1$ is the velocity at $t_1$ and $v_2$ at $t_2$,

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

Two points must be locked in. First, $\Delta v$ is a signed quantity — a change from $+15$ to $+5$ $\text{m s}^{-1}$ gives $\Delta v = -10$, and a change from $+5$ to $-5$ also gives $\Delta v = -10$. Second, on a v–t plot $\bar{a}$ is the slope of the chord joining $(t_1, v_1)$ and $(t_2, v_2)$ — not the slope of a tangent. In one dimension, direction is carried by the algebraic sign; no vector arrow is needed.

Instantaneous acceleration

The instantaneous acceleration is the limit of the average acceleration as the time interval shrinks to zero:

$$a = \lim_{\Delta t \to 0}\frac{\Delta v}{\Delta t} = \frac{dv}{dt} = \frac{d^2x}{dt^2}.$$

The second equality follows because $v = dx/dt$. Geometrically, $a$ at any instant is the slope of the tangent to the v–t curve at that instant. The previous subtopic read $v$ from the slope of x–t; the same logic applies one layer up.

Uniform vs non-uniform acceleration

If $a$ is constant throughout the motion, the acceleration is uniform; $a = \bar{a}$ on every interval and the v–t graph is a straight line. If $a$ varies with time, the acceleration is non-uniform; the v–t graph is curved. The next subtopic — Kinematic equations for uniform acceleration — uses the uniform case to derive the three equations of motion.

Signs of v and a — the four cases that decide speeding up vs slowing down

The most common NEET error in this subtopic is the equation "negative acceleration = deceleration". It is wrong. Deceleration describes a change in speed, not a change in velocity. Whether the body is speeding up or slowing down depends on the relative signs of $v$ and $a$, not on the sign of $a$ alone.

The rule: speed increases when $v$ and $a$ have the same sign, and decreases when they have opposite signs. NCERT Points to Ponder #2: "If a particle is speeding up, acceleration is in the direction of velocity; if its speed is decreasing, acceleration is opposite to velocity." The four panels in NCERT Fig. 2.3 enumerate these cases.

Sign of velocity v Sign of acceleration a Direction of motion What happens to speed Label
$v > 0$ $a > 0$ Along +x axis Speed increases Accelerating
$v > 0$ $a < 0$ Along +x axis Speed decreases Decelerating
$v < 0$ $a < 0$ Along −x axis Speed increases Accelerating
$v < 0$ $a > 0$ Along −x axis Speed decreases Decelerating

Row 3 is the case students miss most often. A stone falling freely (upward = positive) has $v < 0$ and $a = -g < 0$. Both negative — speed is increasing. NCERT Points to Ponder #3: "If a particle is falling under gravity, this acceleration, though negative, results in increase in speed." A shortcut that equates "negative $a$" with "slowing down" misclassifies every falling body.

Reading the v–t graph and the a–t graph

For uniformly accelerated motion the v–t graph is a straight line. Three readings come off it:

  • y-intercept = initial velocity $v_0$ at $t = 0$.
  • Slope at any point = instantaneous acceleration. Straight line $\Rightarrow$ constant $a$.
  • Area under the v–t curve between $t_1$ and $t_2$ = displacement $\Delta x$ over that interval (NCERT extends the rectangle proof to general curves via calculus).

The first two are about $a$ directly; the third is the bridge to the kinematic equations in the next subtopic. For a curved v–t graph (non-uniform $a$), the instantaneous acceleration at any point is the slope of the tangent there.

The a–t graph

The acceleration-time graph plots $a$ vs $t$. For uniform $a$, this is a horizontal line. The area under an a–t graph between $t_1$ and $t_2$ equals $\Delta v$ — the integral counterpart of $a = dv/dt$. NCERT Fig. 2.7(a) shows the a–t graph for free fall: a horizontal line at $a = -g$; the corresponding v–t graph is linear in $t$ with slope $-g$.

Cross-link

The next subtopic uses the v–t-area result above to derive the three kinematic equations for uniform acceleration. Make sure the slope-and-area reading is automatic before you move on.

Worked examples

Worked example 1 · sign discipline on a round trip

A particle on the x-axis has velocity $+12$ $\text{m s}^{-1}$ at $t = 0$ and $-4$ $\text{m s}^{-1}$ at $t = 4$ s. Compute the average acceleration and decide whether the particle is decelerating throughout.

With signed velocities: $\bar{a} = (-4 - 12)/(4 - 0) = -4 \,\text{m s}^{-2}.$

Since $v$ changes from $+12$ to $-4$, the particle reverses direction within the interval. Before the reversal $v > 0$, $a < 0$ (opposite signs) — decelerating. After the reversal $v < 0$, $a < 0$ (same signs) — speeding up. "Decelerating throughout" is wrong; the label flips at the turnaround.

Worked example 2 · instantaneous acceleration from a position polynomial

The position is $x = 4t^3 - 2t^2 + t$ m, $t$ in seconds. Find (i) $v(2)$, (ii) $a(2)$, (iii) $\bar{a}$ over $[0, 2]$ s.

(i) $v(t) = dx/dt = 12t^2 - 4t + 1$. At $t = 2$: $v = 48 - 8 + 1 = 41$ $\text{m s}^{-1}$.

(ii) $a(t) = d^2x/dt^2 = 24t - 4$. At $t = 2$: $a = 48 - 4 = 44$ $\text{m s}^{-2}$.

(iii) $v(0) = 1$ $\text{m s}^{-1}$, $v(2) = 41$ $\text{m s}^{-1}$, so $\bar{a} = (41 - 1)/2 = 20$ $\text{m s}^{-2}$. Note $\bar{a} \neq a(2)$ — when $a$ varies, the average over an interval is generally different from $a$ at the endpoints.

Worked example 3 · acceleration from a piecewise v–t graph

A particle's v–t graph (in $\text{m s}^{-1}$ vs s) is piecewise linear: $v$ rises from $0$ to $20$ over $0 \to 4$ s, stays at $20$ over $4 \to 10$ s, then falls to $0$ over $10 \to 14$ s. Find $a$ in each segment and the total displacement.

Segment 1: slope $= 20/4 = +5$ $\text{m s}^{-2}$. $v > 0$ and $a > 0$ — speeding up.

Segment 2: slope $= 0$. $a = 0$ — uniform velocity (zero $a$ does not mean zero motion).

Segment 3: slope $= -20/4 = -5$ $\text{m s}^{-2}$. $v > 0$, $a < 0$ — decelerating. This is the only segment to which "deceleration" applies.

Total displacement = area under v–t = $\tfrac{1}{2}(4)(20) + (6)(20) + \tfrac{1}{2}(4)(20) = 40 + 120 + 40 = 200$ m.

Quick recap

What this subtopic locked in

  • Definitions. $\bar{a} = \Delta v/\Delta t$; $a = dv/dt = d^2x/dt^2$. SI unit $\text{m s}^{-2}$.
  • Slope of v–t = acceleration. Area under v–t = displacement. Two different readings of the same graph.
  • Deceleration ≠ negative acceleration. Deceleration is a sign-pair condition: $v$ and $a$ opposite. Negative $a$ is just a statement about the axis.
  • Four sign-cases. Same sign of $v$ and $a$ → speeding up; opposite signs → slowing down.
  • Zero velocity does not imply zero acceleration. Ball at the top of a vertical throw still has $a = -g$.
  • Uniform vs non-uniform. Constant $a$ → straight v–t line; varying $a$ → curved v–t line. Kinematic equations apply only to the uniform case.

NEET PYQ Snapshot — Acceleration

Two NEET PYQs that test acceleration ideas — sign handling, average acceleration with reversal, and reading position/velocity from derivatives — plus two NCERT Exercise items that round out the sign-and-graph scope.

NEET 2018

A toy car with charge $q$ moves on a frictionless horizontal plane surface under the influence of a uniform electric field $\vec{E}$. Due to the force $q\vec{E}$ its velocity increases from $0$ to $6$ m/s in one second duration. At that instant the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between $0$ to $3$ seconds are respectively:

  1. $2$ m/s, $4$ m/s
  2. $1$ m/s, $3$ m/s
  3. $1$ m/s, $3.5$ m/s
  4. $1.5$ m/s, $3$ m/s
Answer: (2) 1 m/s, 3 m/s

Why: $a_1 = qE/m = +6$ $\text{m s}^{-2}$ (initial-motion direction positive). Car gains $v = 6$ $\text{m s}^{-1}$ in 1 s. Field reverses $\Rightarrow a_2 = -6$ $\text{m s}^{-2}$: now $v > 0$, $a < 0$ — the decelerating case from our table. Car slows to $0$ at $t = 2$ s, then continues to $v = -6$ $\text{m s}^{-1}$ at $t = 3$ s. Displacement = signed area = $\tfrac{1}{2}(1)(6) + \tfrac{1}{2}(1)(6) - \tfrac{1}{2}(1)(6) = +3$ m, so $\bar{v} = 1$ m/s. Path length = $3 \times \tfrac{1}{2}(1)(6) = 9$ m, so average speed = 3 m/s. Distractor (1) over-counts the displacement to 6 m; (3) mishandles the reversal; (4) uses unsigned area for displacement.

NEET 2016

If the velocity of a particle is $v = At + Bt^2$, where $A$ and $B$ are constants, then the distance travelled by it between $1$ s and $2$ s is:

  1. $3A + 7B$
  2. $\dfrac{3}{2}A + \dfrac{7}{3}B$
  3. $\dfrac{A}{2} + \dfrac{B}{3}$
  4. $\dfrac{3}{2}A + 4B$
Answer: (2)

Why: $a = dv/dt = A + 2Bt$ is time-dependent — non-uniform acceleration. So $x = v_0 t + \tfrac{1}{2}at^2$ does not apply. Use $dx = v\,dt$ and integrate: $x = \int_1^2 (At + Bt^2)\,dt = \left[\tfrac{At^2}{2} + \tfrac{Bt^3}{3}\right]_1^2 = \tfrac{3A}{2} + \tfrac{7B}{3}$. Takeaway: when $a$ varies, go back to definitions and integrate. Distractor (1) forgets to integrate; (3) uses the $t = 1$ endpoint alone; (4) applies $\tfrac{1}{2}at^2$ with $a$ at one instant.

NCERT Exercise 2.7

Read each statement below carefully and state with reasons and examples, if it is true or false. A particle in one-dimensional motion:

  1. with zero speed at an instant may have non-zero acceleration at that instant
  2. with zero speed may have non-zero velocity
  3. with constant speed must have zero acceleration
  4. with positive value of acceleration must be speeding up
Answer: (a) True, (b) False, (c) True for 1-D fixed-direction motion, (d) False

Why: (a) True. Ball at the top of a throw — $v = 0$ but $a = -g$. (b) False. Zero speed means $|v| = 0 \Rightarrow v = 0$. (c) True in 1-D along a fixed direction. False in general 2-D/3-D — uniform circular motion is the counter-example, but here the question restricts to 1-D. (d) False. If $v < 0$ and $a > 0$, the body decelerates despite positive $a$ — row 4 of the sign-case table.

NCERT Exercise 2.18

A speed-time graph of a particle moving along a constant direction is shown with three equal intervals of time. Choosing the positive direction as the constant direction of motion, give the signs of $v$ and $a$ in the three intervals.

Answer: $v > 0$ in all three intervals; $a > 0$ in the first, $a = 0$ in the second, $a < 0$ in the third.

Why: Motion is along the positive direction, so $v > 0$ throughout. Interval 1: speed rising $\Rightarrow$ slope of v–t positive $\Rightarrow a > 0$. Interval 2: speed steady $\Rightarrow a = 0$. Interval 3: speed falling $\Rightarrow a < 0$. Because $v$ stays positive, the sign of $a$ matches the everyday labels (accelerating/decelerating). If $v$ were negative throughout, the labels in intervals 1 and 3 would flip — the row-1-versus-row-3 distinction from our table.

FAQs — Acceleration

Short answers to the questions NEET aspirants ask most about acceleration and its sign conventions.

Is deceleration the same as negative acceleration?
No. Deceleration means the speed (magnitude of velocity) is decreasing. Negative acceleration only means the acceleration vector points in the negative direction of the chosen axis. A body moving in the negative direction with negative acceleration is actually speeding up, not slowing down. The rule: the body decelerates when velocity and acceleration have opposite signs, regardless of which sign is which.
Can a body have zero velocity but non-zero acceleration?
Yes. NCERT Points to Ponder spells this out. A ball thrown straight up has zero velocity at the topmost point, but its acceleration there is still minus g — gravitational acceleration does not pause just because the velocity momentarily reaches zero. Velocity zero at an instant says nothing about the rate of change of velocity at that instant.
What does the slope of the velocity-time graph tell you?
The slope of the v–t graph at any instant gives the instantaneous acceleration at that instant. A straight inclined line means constant acceleration. A horizontal line means zero acceleration (uniform velocity). A curved v–t graph means the acceleration itself is changing — non-uniform acceleration. The area under the v–t curve, separately, equals the displacement over that interval.
If speed is constant, must acceleration be zero?
In one-dimensional motion along a fixed straight line, yes — constant speed plus unchanging direction means constant velocity, so acceleration is zero. The statement is false in general two or three-dimensional motion: a body in uniform circular motion has constant speed yet non-zero acceleration because the direction of velocity changes. NCERT Exercise 2.7(c) tests this in the 1-D form, where the answer is "true".
How do you find instantaneous acceleration from a position function x(t)?
Differentiate twice. The first derivative dx/dt gives the velocity v(t); differentiating once more gives the acceleration a(t) = d²x/dt². For x = 4t³ − 2t² + t, v = 12t² − 4t + 1 and a = 24t − 4 — measured in m/s² when t is in seconds. The same logic applies if you start from v(t): a(t) = dv/dt directly.
Does positive acceleration always mean a body is speeding up?
No. Positive acceleration means the acceleration vector points along the positive direction of the chosen axis. Whether the body speeds up depends on the velocity. If velocity is positive and acceleration is positive, the body speeds up. But if velocity is negative and acceleration is positive, the body slows down — it is decelerating despite the positive value of a. The sign-pair, not the sign alone, decides.
Related Topics
Motion in a Straight Line — Parent chapter Full chapter overview: position, velocity, acceleration, kinematic equations, relative velocity, free fall, PYQs. Position, Path Length & Displacement Reference frames, scalar path length vs vector displacement, sign conventions on a number line. Average vs Instantaneous Velocity Average velocity, harmonic-mean speed trap, instantaneous velocity as the slope of x–t. Kinematic Equations for Uniform Acceleration v = v₀ + at, x = v₀t + ½at², v² = v₀² + 2ax — derivations and NEET-style problems. Relative Velocity in One Dimension Same-direction and opposite-direction relative velocity, train and bullet problems. Motion Under Gravity (Free Fall) Vertical throws, height of fall, Galileo's odd-numbers law, NEET-classic projectile-up problems.