Acceleration in Straight Line Motion
Velocity tells you how fast position changes; acceleration tells you how fast velocity itself changes — the rate of a rate. It is the concept that turns descriptive kinematics into predictive mechanics, because in the next chapter forces will step in and cause acceleration (F = ma). Within this chapter, acceleration is what makes the kinematic equations possible: they work precisely when a stays constant. NTA's favourite testing angle is the sign logic — candidates who equate "negative acceleration" with "slowing down" lose the mark every single year — along with v–t slope readings and short average-acceleration numericals. This page defines both average and instantaneous acceleration, builds the sign logic rigorously, and introduces the two graph skills — slope and area — that the rest of the chapter runs on.
What is Acceleration? — Complete Theory
1. Average and instantaneous definitions
If velocity changes from v₁ to v₂ over the interval Δt, the average acceleration is
ā = Δv/Δt = (v₂ − v₁)/Δt — SI unit m s⁻²; dimensions [L T⁻²] (velocity [L T⁻¹] per time [T]).
The instantaneous acceleration is the limit of this as Δt → 0, i.e. the derivative of the velocity function:
a = dv/dt = d²x/dt² — the second derivative of position.
Both carry a sign in 1D: a = −2 m s⁻² is an acceleration of magnitude 2 m s⁻² along the −x direction. Note carefully what acceleration is not: it is not the "amount of motion" (that is velocity) and not the "amount of ground covered" (that is distance). A jet cruising at 900 km h⁻¹ in a straight line has zero acceleration; a ball at the instant it is released from rest has 9.8 m s⁻² of it. Always dimension-check derived accelerations: differentiating v in m s⁻¹ against t in s must yield m s⁻² — if it does not, a differentiation slip has crept in.
2. The sign logic: speeding up vs slowing down
The direction of acceleration relative to velocity — not the sign of acceleration alone — decides whether speed grows or shrinks. In 1D the test is the sign product v·a:
| v | a | v·a | Result | Example |
|---|---|---|---|---|
| +5 m s⁻¹ | +2 m s⁻² | > 0 | speeding up | car pressing the accelerator |
| +5 m s⁻¹ | −2 m s⁻² | < 0 | slowing down (deceleration) | car braking |
| −5 m s⁻¹ | −2 m s⁻² | > 0 | speeding up (in −x) | ball falling with down = negative |
| −5 m s⁻¹ | +2 m s⁻² | < 0 | slowing down | ball rising, up = positive |
Terminology: deceleration (or retardation) is not "negative acceleration" — it means the speed is decreasing, i.e. a opposes v regardless of signs. This is the single most examined subtlety of the topic, and it also explains why a body at the top of a vertical throw (v = 0) still has full gravitational acceleration: the velocity is about to become negative while a stays −g throughout (the free fall page works this example in full).
3. Two graph skills: slope and area on the v–t plot
The velocity–time graph is the workbench of this chapter, and it supports exactly two operations. Slope gives acceleration: the gradient of the v–t curve at any instant is dv/dt — a straight slanted line means constant a, a horizontal line means a = 0, and a curved line means a itself is changing (not zero!). Area gives displacement: since [v]·[t] = [L], the area trapped between the v–t curve and the time axis equals the displacement in that interval; areas above the axis count positive, below it negative. The area skill is justified properly by integration on the kinematic equations page and becomes the standard way to handle piecewise motion in JEE Main numericals. Master the pair — slope up, area down — and most graph questions reduce to two look-ups.
Visualising Acceleration: the v–t Story
Figure 2.4 — Three phases of one-dimensional motion
Read left to right: the rising segment has positive slope (a > 0, speed growing); the flat segment has zero slope (a = 0, velocity constant); the falling segment has negative slope (a < 0, speed shrinking while v stays positive). The whole story stays above the t-axis, so the velocity never changed direction.
Exam read-out: slope is read vertically (steeper → larger |a|); area is read down to the axis (taller plateau → more displacement). If the falling segment continued below the t-axis, the velocity would have reversed and the area below would count negative — the standard bridge to braking-then-reversing problems in JEE Main.
Solved Examples on Acceleration (Step-by-Step)
Solved Example 1 — Constant braking, read two ways
Q. A particle's velocity is v = 20 − 5t (SI). Find (i) the acceleration, (ii) when it is momentarily at rest, and (iii) how it moves just after that instant.
Step 1. a = dv/dt = −5 m s⁻² — constant (v is linear in t; the v–t graph is a straight falling line).
Step 2. Rest: v = 0 → 20 − 5t = 0 → t = 4 s.
Step 3. For t > 4 s, v turns negative while a stays −5: v·a > 0 → the particle speeds up in the −x direction. The braking phase (0 to 4 s, v·a < 0) hands over to a reverse acceleration phase.
ANSWER: (i) −5 m s⁻² throughout; (ii) at rest at t = 4 s; (iii) speeds up along −x after t = 4 sSolved Example 2 — Average acceleration with a direction flip
Q. During a slow-motion replay, a ball's velocity changes from 10 m s⁻¹ downward to 10 m s⁻¹ upward in Δt = 4 s. Taking upward as positive, find the average acceleration over this interval.
Step 1. Δv = v₂ − v₁ = (+10) − (−10) = +20 m s⁻¹ — the reversal doubles the change.
Step 2. ā = Δv/Δt = 20/4 = +5 m s⁻², i.e. 5 m s⁻² upward.
Step 3. Trap check: naively subtracting magnitudes (10 − 10 = 0) would miss the direction change entirely — always convert to signed velocities first.
ANSWER: ā = +5 m s⁻² (upward)Practice Questions on Acceleration (With Solutions)
Practice Q1 (Numerical-value type)
Q. A particle's position is x = 4t³ − 6t (SI). Its acceleration at t = 2 s is ______ m s⁻².
Solution. v = dx/dt = 12t² − 6; a = dv/dt = 24t. At t = 2 s: a = 48 m s⁻². (Second derivative of position, power rule twice.)
ANSWER: 48Practice Q2 (Single-correct MCQ)
Q. A body moving along +x has v = −3 m s⁻¹ and a = −1 m s⁻². It is: (a) slowing down (b) speeding up (c) at rest (d) moving with constant speed
Solution. v·a = (−3)(−1) = +3 > 0 → velocity and acceleration point the same way → speed is growing (from 3 m s⁻¹ onward). Negative signs describe direction, not decrease.
ANSWER: (b)Practice Q3 (Assertion–Reason)
Q. Assertion (A): A body can have zero velocity and yet non-zero acceleration. Reason (R): Acceleration is the rate of change of velocity, so a momentarily-zero velocity can still be changing.
Options: (a) both true, R explains A · (b) both true, R does not explain A · (c) A true, R false · (d) A false, R true
Solution. The top of a vertical throw: v = 0 at the crest while a = 9.8 m s⁻² downward — the velocity is passing through zero, not staying there. R states exactly why A is possible.
ANSWER: (a)Key Formulas & Takeaways
| Rule | Statement | Fast example |
|---|---|---|
| Average acceleration | ā = Δv/Δt (signed) | +10 → −10 in 4 s → +5 m s⁻² |
| Instantaneous | a = dv/dt = d²x/dt² | x = 4t³ → a = 24t |
| Speed test | v·a > 0 speeding up; v·a < 0 slowing | v = −5, a = −2 → speeding up |
| v–t slope | gradient = acceleration | straight line → constant a |
| v–t area | area between curve and t-axis = displacement (signed) | plateau 10 × 4 s → 40 m |
| Units check | [a] = [L T⁻²]; SI m s⁻² | g = 9.8 m s⁻² ✔ |
| Constant-a shortcut | kinematic equations (next page) | valid only when a is fixed |
FAQs on Acceleration
What is acceleration in one-dimensional motion?
It is the rate of change of velocity: a = dv/dt = d²x/dt², with SI unit m s⁻². A positive acceleration means the velocity is becoming more positive, not that the object is speeding up.
Does negative acceleration always mean slowing down?
No. The body slows only when acceleration opposes velocity (v·a < 0). If both are negative — for example v = −5 m s⁻¹ and a = −2 m s⁻² — the speed grows from 5 to 7 m s⁻¹, so the body speeds up.
What does the area under a velocity-time graph represent?
Displacement: velocity × time has dimensions of length. Regions below the time axis count as negative, so total distance requires the areas to be counted without their signs.
Can a body have zero velocity and non-zero acceleration?
Yes — at the top of a vertical throw the velocity is zero while the acceleration is 9.8 m s⁻² downward. Zero velocity means momentarily at rest, not no forces acting.
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