A car moving at 30 m/s applies brakes that produce a uniform deceleration of 6 m/s². Using the kinematic equat

Example · Example 5

Q. A car moving at 30 m/s applies brakes that produce a uniform deceleration of 6 m/s². Using the kinematic equations, find the time taken by the car to stop and the distance it travels before stopping.

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Concept understanding — Uniformly Accelerated Motion

Uniformly Accelerated Motion

Imagine you're sitting in a train that starts moving from a station. At first, it crawls — then it picks up speed smoothly, second by second. If you watch the speedometer, you might see it climb by the same amount every second: 0 to 10 km/h, then 10 to 20, then 20 to 30. That steady, predictable increase is the heart of uniformly accelerated motion.

The Intuition

When something moves with uniform acceleration, its velocity changes by the same amount in every equal interval of time. The change is constant — not faster one second and slower the next.

Think of a ball rolling down a gentle, straight ramp. It starts from rest. In the first second, it gains some speed. In the next second, it gains exactly the same amount of speed again. The acceleration — the rate of change of velocity — is fixed.

Note

"Uniform" here means "constant" or "unchanging." It does not mean the speed is constant. In fact, the speed is changing — but the rate at which it changes is constant.

The Precise Statement

Uniformly accelerated motion is motion in a straight line where the acceleration $a$ is constant in both magnitude and direction.

Mathematically, if $v$ is velocity at time $t$ , and $u$ is the initial velocity (at $t = 0$ ), then:

a = \frac{v - u}{t} = \text{constant}

This single idea leads to the three famous equations of motion (for constant acceleration):

v = u + at

Here:

$u$ = initial velocity (at $t = 0$ )
$v$ = velocity at time $t$
$a$ = constant acceleration
$s$ = displacement in time $t$

What It Looks Like in Real Life

Situation	Acceleration	Why it's (approximately) uniform
A car accelerating on a highway	~2–3 m/s²	Engine provides roughly constant force
A ball dropped from a height	9.8 m/s² downward	Gravity is nearly constant near Earth's surface
A train starting from a station	~0.5 m/s²	Controlled by the driver to be smooth

Watch out

Not all motion is uniformly accelerated. A car stopping suddenly has deceleration that changes — it's not uniform. A roller coaster has acceleration that varies wildly. Uniform acceleration is an ideal model that works beautifully for many real situations (like free fall) but not all.

The Key Insight

The word "uniform" refers to the acceleration, not the velocity. If acceleration is constant, then:

Velocity changes linearly with time (a straight line on a $v$ - $t$ graph)
Displacement changes quadratically with time (a parabola on an $s$ - $t$ graph)

This is why the equations above are so powerful: they let you predict position and velocity at any instant, as long as acceleration stays constant.

Important

For uniformly accelerated motion, the $v$ - $t$ graph is always a straight line. The slope of that line equals the acceleration. If the graph is curved, acceleration is not uniform. …

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Setting up. Take the car's initial direction of motion as positive. Initial velocity $u = 30$ m/s, final velocity $v = 0$ (the car stops), and the deceleration has magnitude $6\ \text{m/s}^2$ , so the acceleration is $a = -6\ \text{m/s}^2$ (negative because it opposes the velocity — see Section 2.5 on retardation).

Time to stop. Using $v = u + at$ (Section 2.8): $0 = 30 + (-6)t \quad\Longrightarrow\quad t = \frac{30}{6} = 5\ \text{s}$
Stopping distance. Using $v^2 = u^2 + 2as$ : $0 = (30)^2 + 2(-6)s \quad\Longrightarrow\quad 0 = 900 - 12s \quad\Longrightarrow\quad s = \frac{900}{12} = 75\ \text{m}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set SET-AP55001 mark
$Q. The rate of change of velocity of an object is called ________.$
›Reveal solutionSolution

The rate of change of velocity of an object with respect to time is called its acceleration.

Velocity can change in magnitude (speeding up/slowing down), direction, or both. Acceleration is defined as:

a = dv/dt = (change in velocity)/(time taken)

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Velocity of an object moving with uniform acceleration will be:$
(a) decreasing
(b) increasing
(c) will be zero
(d) increasing or decreasing
›Reveal solutionSolution

With uniform acceleration the velocity may either increase or decrease.

Uniform acceleration means the velocity changes by equal amounts in equal times. If the acceleration is in the same direction as the velocity, the speed increases; if it is opposite (retardation), the speed decreases.

…

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CBSE 2022Set ANNUAL1 markMCQ
$Q. An object is moving with a constant acceleration. Its velocity time graph is:$
(a) graph A
(b) graph B
(c) graph C
(d) graph D
›Reveal solutionSolution
Constant acceleration means a straight-line v–t graph with constant slope — that's graph (A).

Reasoning. For constant acceleration $a$ , the equation of motion is $v = u + at$ , which is linear in $t$ (a straight line with slope $a$ and intercept $u$ ).

Checking each sketch:

(A) Straight line from the origin with constant positive slope — velocity increases uniformly with time, exactly $v=at$ (starting from rest). This matches constant acceleration.

(B) Horizontal line — velocity doesn't change with time, so acceleration is zero, not constant nonzero acceleration.

(C) A vertical line — implies velocity changing instantaneously with no time elapsed, i.e. infinite acceleration, which is unphysical. …
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CBSE 2022Set ANNUAL1 markMCQ
$Q. If the slope of the velocity time graph of a particle moving in stright line be constant, then its motion is with —$
(a) Uniform velocity
(b) Uniform acceleration
(c) Variable acceleration
(d) Constant speed
›Reveal solutionSolution

Constant slope of a v–t graph ⇒ uniform acceleration.

Acceleration $a = \dfrac{dv}{dt}$ = slope of the velocity–time graph.

…

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CBSE 2021Set ANNUAL1 markMCQ
$Q. The distance travelled by a particle is directly proportional to the square of the time taken, the acceleration of the particle is :$
(a) Increasing
(b) Decreasing
(c) Zero
(d) Constant
›Reveal solutionSolution

$s = kt^2$ differentiates to a constant acceleration $a = 2k$ .

Given $s \propto t^2$ , write $s = k t^2$ for some constant $k$ .

Velocity: $v = \dfrac{ds}{dt} = 2kt$

Acceleration: $a = \dfrac{dv}{dt} = 2k$

…

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CBSE 2019Set ANNUAL1 markMCQ
$Q. Which graph represents uniform acceleration ?$
(a) graph
(a) -- straight line through origin, constant slope
(b) graph
(b) -- triangular rise-then-fall
(c) graph
(c) -- concave-up increasingly steep curve
(d) graph
(d) -- concave-down flattening/saturating curve
›Reveal solutionSolution

A displacement-time graph for uniform acceleration is a concave-up curve that gets steeper with time, since s depends on t^2 -- graph (c).

For motion with uniform (constant) acceleration a, the displacement-time relation is

s = ut + (1/2)a t^2

This is a quadratic function of time. Its slope, ds/dt = u + at, is the instantaneous velocity, which increases steadily with time for positive a. A steadily increasing slope means the curve bends upward more and more sharply as t increases -- this is exactly a concave-up curve that starts at the origin and curves increasingly steeply, matching graph (c).

By contrast:

(a) a straight line through the origin with constant slope describes UNIFORM VELOCITY (zero acceleration), not uniform acceleration. …

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Q.A car moving at 30 m/s applies brakes that produce a uniform deceleration of 6 m/s². Using the kinematic equations, find the time taken by the car to stop and the distance it travels before stopping.

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