Three equations. Five quantities. When acceleration is constant, any three of the five (u, v, a, s, t) can unlock the other two. From NCERT Chapter 4 (Exploration edition) Class 9 Science. Aligned with CBSE syllabus 2026-27.
When an object moves with constant acceleration in a straight line, its motion is completely described by five quantities. Knowing any three of them lets you calculate the remaining two using the kinematic equations.
| Symbol | Name | SI Unit | Scalar / Vector |
|---|---|---|---|
| u | Initial velocity | m s⁻¹ | Vector |
| v | Final velocity | m s⁻¹ | Vector |
| a | Acceleration | m s⁻² | Vector |
| s | Displacement | m | Vector |
| t | Time interval | s | Scalar |
You could memorise the three equations — but understanding where they come from is far more valuable. Each equation is derived from fundamental definitions (acceleration = change in velocity ÷ time; displacement = area under v-t graph). Knowing the derivation means you never misremember a formula — you can re-derive it in 30 seconds. It also tells you exactly when the equations are valid and when they are not.
From the definition of average acceleration (NCERT Eq. 4.3c):
$$a = \frac{v - u}{t}$$Multiply both sides by $t$:
$$at = v - u$$Rearrange:
$$\boxed{v = u + at} \quad \cdots \text{(Eq. 4.4a)}$$Final velocity = Initial velocity + (Acceleration × Time)
Each second of acceleration adds $a$ m s⁻¹ to the initial velocity. If $a$ is positive, $v > u$ (speeding up). If $a$ is negative, $v < u$ (slowing down).
Missing quantity: This equation does not contain displacement $s$. Use it when $s$ is not needed or not given.
On a velocity-time graph for constant acceleration: the y-intercept (value at $t = 0$) equals $u$, and the slope equals $a$. Reading off the graph at time $t$ gives exactly $v = u + at$. The equation is the algebraic statement of "slope × time + initial value."
Consider the velocity-time graph for an object with initial velocity $u$ and constant acceleration $a$ over time interval $t$. From Eq. 4.4a, the final velocity is $v = u + at$.
The area under this graph (a trapezium OABD in Fig. 4.19) equals the displacement $s$. Split the trapezium into:
Total displacement:
$$s = ut + \frac{1}{2}at^2$$ $$\boxed{s = ut + \frac{1}{2}at^2} \quad \cdots \text{(Eq. 4.4b)}$$Alternatively, use the average velocity approach. For constant acceleration, average velocity = (u + v)/2. Displacement = average velocity × time:
$$s = \frac{u + v}{2} \times t$$Substitute $v = u + at$ from Eq. 4.4a:
$$s = \frac{u + (u + at)}{2} \times t = \frac{2u + at}{2} \times t = ut + \frac{1}{2}at^2$$Displacement = (initial velocity × time) + (½ × acceleration × time²)
The first term $ut$ is the displacement the object would have covered even if it kept moving at its initial velocity. The second term $\frac{1}{2}at^2$ is the extra displacement due to acceleration. They add up to give the total.
Special case: If $u = 0$ (starts from rest), $s = \frac{1}{2}at^2$ — displacement grows with the square of time. Double the time → four times the displacement.
Missing quantity: This equation does not contain final velocity $v$.
From Eq. 4.4a: $v = u + at \Rightarrow t = \dfrac{v - u}{a}$
Substitute into Eq. 4.4b: $s = ut + \frac{1}{2}at^2$
$$s = u \cdot \frac{v-u}{a} + \frac{1}{2}a \cdot \left(\frac{v-u}{a}\right)^2$$ $$s = \frac{u(v-u)}{a} + \frac{1}{2} \cdot \frac{(v-u)^2}{a}$$ $$as = u(v-u) + \frac{1}{2}(v-u)^2$$ $$2as = 2u(v-u) + (v-u)^2$$ $$2as = 2uv - 2u^2 + v^2 - 2uv + u^2$$ $$2as = v^2 - u^2$$ $$\boxed{v^2 = u^2 + 2as} \quad \cdots \text{(Eq. 4.4c)}$$This equation connects velocity directly to displacement — without needing time.
It is especially useful when the time taken is not given and not asked for — you can find the stopping distance of a braking car, or the velocity after sliding a certain distance, directly.
Special case: If the object stops ($v = 0$): $0 = u^2 + 2as \Rightarrow s = -\dfrac{u^2}{2a}$. The negative sign on $a$ (retardation) makes $s$ positive.
Missing quantity: This equation does not contain time $t$.
NCERT's "Journey Beyond" section introduces two additional useful equations, both derivable from the three main ones:
Derivation: From Eq. 4.4a, $u = v - at$. Substitute into Eq. 4.4b:
$$s = (v - at)t + \frac{1}{2}at^2 = vt - at^2 + \frac{1}{2}at^2 = vt - \frac{1}{2}at^2$$Use: When final velocity $v$ and time $t$ are known, but initial velocity $u$ is not. It is the "backwards" version of Eq. 4.4b.
Derivation: This is the trapezium area formula directly. Displacement = average velocity × time, and for constant acceleration, average velocity = (u + v)/2:
$$s = \frac{u + v}{2} \times t = \frac{1}{2}(u + v)t$$Use: When both initial and final velocities are known and time is known. It bypasses acceleration entirely.
All five equations at a glance:
The kinematic equations are valid only when ALL three conditions below are satisfied:
Do not use kinematic equations for: circular motion, projectile motion (without resolving), or any motion with changing acceleration (e.g., car in real traffic).
The most common mistake students make is using the wrong equation. Here is a decision guide:
| Given quantities | Unknown | Use equation |
|---|---|---|
| u, a, t | v | v = u + at |
| u, a, t | s | s = ut + ½at² |
| u, v, a | s | v² = u² + 2as |
| u, v, s | a | v² = u² + 2as |
| u, v, t | s | s = ½(u + v)t |
| u, v, t | a | v = u + at → solve for a |
| u = 0, a, t | s | s = ½at² |
| u = 0, a, s | v | v² = 2as |
A car is moving at (a) 54 km h⁻¹ and (b) 108 km h⁻¹. The brakes produce a constant retardation of 4 m s⁻². Find the stopping distance in each case.
Convert speeds: 54 km h⁻¹ = 54 × 5/18 = 15 m s⁻¹; 108 km h⁻¹ = 30 m s⁻¹
Given: $v = 0$ (stops), $a = -4$ m s⁻² (retardation). Unknown: $s$. Use $v^2 = u^2 + 2as$.
(a) u = 15 m s⁻¹:
$$0 = 15^2 + 2(-4)s \Rightarrow 8s = 225 \Rightarrow s = \mathbf{28.125 \approx 28.1 \text{ m}}$$(b) u = 30 m s⁻¹:
$$0 = 30^2 + 2(-4)s \Rightarrow 8s = 900 \Rightarrow s = \mathbf{112.5 \text{ m}}$$Key conclusion: Doubling the speed quadruples the stopping distance (225 → 900; ratio 4:1). This has critical implications for road safety — at highway speeds, braking distances become dangerously long.
A car starts from rest and attains a velocity of 24 m s⁻¹ in 6 s. Find (a) the acceleration and (b) the distance covered.
Given: $u = 0$, $v = 24$ m s⁻¹, $t = 6$ s.
(a) Acceleration: $a = (v-u)/t = (24-0)/6 = \mathbf{4 \text{ m s}^{-2}}$
(b) Distance: Use $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(4)(36) = \mathbf{72 \text{ m}}$
Verify with Eq. 3: $v^2 = 2as \Rightarrow 576 = 2(4)s \Rightarrow s = 72$ m ✓
A motorbike moving at 28 m s⁻¹ decelerates and covers 98 m before stopping. Find (a) deceleration and (b) time taken to stop.
Given: $u = 28$ m s⁻¹, $v = 0$, $s = 98$ m.
(a) Use $v^2 = u^2 + 2as$:
$$0 = 784 + 2a(98) \Rightarrow 196a = -784 \Rightarrow a = \mathbf{-4 \text{ m s}^{-2}}$$Deceleration = 4 m s⁻² (retardation of 4 m s⁻²)
(b) Use $v = u + at$:
$$0 = 28 + (-4)t \Rightarrow t = \mathbf{7 \text{ s}}$$A truck slows from 54 km h⁻¹ to 36 km h⁻¹ in 36 s. Find the distance covered during this time.
Convert: $u = 54 \times 5/18 = 15$ m s⁻¹; $v = 36 \times 5/18 = 10$ m s⁻¹; $t = 36$ s.
Use $s = \frac{1}{2}(u + v)t$:
$$s = \frac{1}{2}(15 + 10)(36) = \frac{1}{2}(25)(36) = \mathbf{450 \text{ m}}$$Verify via $a$: $a = (10-15)/36 = -5/36$ m s⁻²; $s = 15(36) + \frac{1}{2}(-5/36)(36)^2 = 540 - 90 = 450$ m ✓
A bus is travelling at 36 km h⁻¹. The driver sees an obstacle 30 m ahead and applies brakes after a reaction time of 0.5 s. The braking produces a deceleration of 8 m s⁻². Will the bus hit the obstacle?
Convert: $u = 36 \times 5/18 = 10$ m s⁻¹; obstacle at 30 m.
Phase 1 — Reaction time (brakes not yet applied, constant velocity):
$$s_1 = u \times t_{reaction} = 10 \times 0.5 = 5 \text{ m}$$Phase 2 — Braking (a = −8 m s⁻², u = 10 m s⁻¹, v = 0):
$$v^2 = u^2 + 2as_2 \Rightarrow 0 = 100 + 2(-8)s_2 \Rightarrow s_2 = \frac{100}{16} = 6.25 \text{ m}$$Total distance = 5 + 6.25 = 11.25 m
Since 11.25 m < 30 m, the bus stops well before the obstacle. It will NOT hit the obstacle.
Safety margin = 30 − 11.25 = 18.75 m. The obstacle is safe.
N1 (Easy). A ball is thrown upward with a velocity of 19.6 m s⁻¹. Taking $g = 9.8$ m s⁻² downward, how long does it take to reach the maximum height?
At maximum height, $v = 0$. Taking upward as positive: $u = 19.6$ m s⁻¹, $a = -9.8$ m s⁻²
$$v = u + at \Rightarrow 0 = 19.6 - 9.8t \Rightarrow t = \mathbf{2 \text{ s}}$$N2 (Easy). A stone dropped from a cliff reaches the ground in 5 s. Find the height of the cliff. ($g = 9.8$ m s⁻²)
$u = 0$, $a = 9.8$ m s⁻², $t = 5$ s
$$s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(9.8)(25) = \mathbf{122.5 \text{ m}}$$N3 (Easy). A cyclist accelerates from rest at 0.4 m s⁻² for 20 s. Find (a) final velocity and (b) distance covered.
(a) $v = 0 + 0.4 \times 20 = \mathbf{8 \text{ m s}^{-1}}$
(b) $s = 0 + \frac{1}{2}(0.4)(400) = \mathbf{80 \text{ m}}$
N4 (Medium). A car moving at 20 m s⁻¹ sees a dog 50 m ahead. The brakes produce deceleration of 4 m s⁻². (a) Will it stop in time? (b) If not, what is the impact speed?
Use $v^2 = u^2 + 2as$ with $u = 20$, $a = -4$, $v = 0$ to find stopping distance:
$$0 = 400 - 8s \Rightarrow s = 50 \text{ m}$$Stopping distance exactly equals 50 m — the car just barely stops at the dog's position. (In a real scenario, since $s = 50$ m exactly, the car stops with zero velocity right at the obstacle.)
N5 (Medium). An object moving with velocity 8 m s⁻¹ decelerates at 2 m s⁻² and travels a distance of 12 m. Find the velocity at that point.
N6 (Medium). A train covers the first 10 km in 4 min, the next 10 km in 5 min. Find the average speed for the 20 km journey in km h⁻¹.
Total time = 4 + 5 = 9 min = 9/60 h. Total distance = 20 km.
$$\text{Average speed} = \frac{20}{9/60} = 20 \times \frac{60}{9} = \mathbf{133.3 \text{ km h}^{-1}}$$N7 (Medium). A bullet is fired horizontally with velocity 400 m s⁻¹ and travels 1000 m before hitting a target. Find the time of travel. (Assume no gravity for horizontal motion.)
Uniform motion horizontally ($a = 0$): $s = ut + \frac{1}{2}at^2 = ut$
$$t = s/u = 1000/400 = \mathbf{2.5 \text{ s}}$$N8 (Hard). Two cars A and B are 100 m apart, moving towards each other. A moves at 15 m s⁻¹ and decelerates at 1 m s⁻². B moves at 10 m s⁻¹ at constant speed. Do they collide?
Car A stops after time $t_A = v/a = 15/1 = 15$ s, covering $s_A = (15)^2/(2 \times 1) = 112.5$ m.
In the same 15 s, car B covers $s_B = 10 \times 15 = 150$ m.
Together they would need to cover 100 m to meet. $s_A + s_B = 112.5 + 150 = 262.5$ m, but they only need to close 100 m. Yes, they will collide. To find when: set $s_A(t) + s_B(t) = 100$: $(15t - 0.5t^2) + 10t = 100$ → $25t - 0.5t^2 = 100$ → $t^2 - 50t + 200 = 0$ → $t = (50 - \sqrt{2500-800})/2 = (50 - \sqrt{1700})/2 \approx 4.3$ s.
N9 (Hard). A body starts from rest and travels 9 m in the 3rd second of its motion. Find the acceleration. (Use: distance in nth second = $u + \frac{a}{2}(2n-1)$)
Using the formula for distance in nth second with $u = 0$, $n = 3$:
$$s_n = u + \frac{a}{2}(2n-1) = 0 + \frac{a}{2}(2 \times 3 - 1) = \frac{5a}{2}$$ $$\frac{5a}{2} = 9 \Rightarrow a = \mathbf{3.6 \text{ m s}^{-2}}$$N10 (Hard). A stone is thrown vertically upward with a velocity of 29.4 m s⁻¹. Find (a) maximum height, (b) time to return to the starting point, (c) velocity when it returns. ($g = 9.8$ m s⁻²)
Take upward as positive. $u = 29.4$ m s⁻¹, $a = -9.8$ m s⁻².
(a) Maximum height ($v = 0$ at top):
$$v^2 = u^2 + 2as \Rightarrow 0 = (29.4)^2 - 2(9.8)H \Rightarrow H = \frac{864.36}{19.6} = \mathbf{44.1 \text{ m}}$$(b) Time to rise: $v = u + at \Rightarrow 0 = 29.4 - 9.8t \Rightarrow t_{up} = 3$ s. By symmetry, total time = $\mathbf{6 \text{ s}}$.
(c) Velocity on return: By symmetry, the stone returns with speed 29.4 m s⁻¹ downward = −29.4 m s⁻¹ (downward).