A cyclist starts from the centre O of a circular park of radius $1\ \text{km}$ and rides along the path O$\to$P$\to$R$\to$Q$\to$O. Here P is a point on the boundary circle reached by riding straight out along a radius (due east of O); the cyclist then rides along the circular boundary from P to R (R lies on the circle midway, in the north-east direction) and on to Q (the point on the circle due north of O); finally the cyclist returns straight along the radius QO back to the centre. Throughout the ride the speed is a constant $10\ \text{ms}^{-1}$. Find the magnitude and direction of the cyclist's acceleration at point R.

A cyclist starts from the centre O of a circular park of radius 1km and rides along the path O→P→R→Q→O. Here P

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1\ \text{km}

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Concept understanding — Uniform Circular Motion

Uniform Circular Motion

Uniform circular motion is motion along a circular path at constant speed. Although the speed stays the same, the velocity does not — its direction keeps changing at every instant — so the motion has an acceleration even though the speed never changes. This concept covers the kinematics of that motion; the force that causes it (centripetal force) belongs to the Laws of Motion unit.

1. Angular Quantities

As the body sweeps through an angle $\theta$ , its angular velocity is

\omega = \frac{d\theta}{dt}

For one full revolution in a period $T$ , at frequency $f$ :

\omega = \frac{2\pi}{T} = 2\pi f, \qquad T = \frac{1}{f}

To convert rpm to rad/s:

\omega = \frac{2\pi \cdot (\text{rpm})}{60} = \frac{\pi}{30}\,(\text{rpm})

The linear (rim) speed $v$ and the angular speed $\omega$ are related by

v = \omega r

2. Centripetal Acceleration

Even though the speed is constant, the velocity vector keeps turning — this produces an acceleration directed toward the centre of the circle:

a_c = \frac{v^2}{r} = \omega^2 r = \frac{4\pi^2 r}{T^2} = 4\pi^2 f^2 r

Use whichever form matches the data you are given. This acceleration is often expressed as a multiple of $g$ (as $a_c/g$ , taking $g = 10\ \text{m/s}^2$ ).

3. Directions

The velocity is always tangential (along the direction of motion); the acceleration is centripetal — radial, pointing inward, and perpendicular to the velocity.

Because the direction of the velocity keeps changing, the change in the velocity vector over an angle $\Delta\theta$ has magnitude

|\Delta v| = 2v\sin\left(\frac{\Delta\theta}{2}\right)

So a quarter turn gives $|\Delta v| = v\sqrt{2}$ , a half turn gives $2v$ , and a full turn gives $0$ .

The average acceleration over an arc is $|\Delta v|/\Delta t$ — this is smaller in magnitude than the instantaneous centripetal acceleration $v^2/r$ , and its direction lies along the perpendicular bisector of the chord joining the two points (which, for a circle, always passes through the centre) rather than being radially inward from either endpoint's own position.

4. Points on a Rotating Body

Every point on a rigid rotating body (a wheel, a disc, a clock hand, the Earth) shares the same $\omega$ , but the rim speed $v = \omega r$ grows with the radius.

Clock hands: the second hand turns at $\omega = \dfrac{2\pi}{60}$ rad/s, the minute hand at $\dfrac{2\pi}{3600}$ rad/s, the hour hand at $\dfrac{2\pi}{43200}$ rad/s.
The Earth spins with $\omega = \dfrac{2\pi}{86400} \approx 7.3\times10^{-5}$ rad/s. A point on the equator moves at $\omega R \approx 465$ m/s, and a point at latitude $\lambda$ moves at $\omega R\cos\lambda$ (since the circle of latitude has radius $R\cos\lambda$ ).

5. Connected Systems

Wheels joined by a belt (or two gears in mesh) share the same rim speed, so $\omega_1 r_1 = \omega_2 r_2$ — the larger wheel turns with the smaller $\omega$ .
Wheels on the same axle (concentric) share the same $\omega$ , so the outer rim moves faster ( $v \propto r$ ).

6. Non-Uniform Circular Motion

If the speed also changes, there is a tangential acceleration

a_t = \frac{dv}{dt} = r\alpha

(from the angular acceleration $\alpha = d\omega/dt$ ), in addition to the radial $a_c = v^2/r$ . These two are perpendicular, so the total acceleration is

$a = \sqrt{a_c^2 + a_t^2}, \qquad \tan\beta = \frac{a_t}{a_c}$ …

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At R the cyclist is on the circular part of the path moving at constant speed, so there is no tangential acceleration — only centripetal acceleration toward the centre. Its magnitude is $a=v^2/r = 0.1\ \text{ms}^{-2}$ and it points along RO (from R to O).

Concept

Acceleration has two parts: a tangential part $a_t = dv/dt$ that changes the speed, and a centripetal part $a_c = v^2/r$ that changes the direction and always points toward the centre of the circular path.

Steps

The speed is constant ( $10\ \text{ms}^{-1}$ ) everywhere, so $a_t = dv/dt = 0$ .
Point R lies on the circular boundary (radius $r = 1\ \text{km} = 1000\ \text{m}$ ), so the motion there is along a circle and the centripetal term is present: $a = a_c = \frac{v^2}{r} = \frac{(10\ \text{ms}^{-1})^2}{1000\ \text{m}} = \frac{100}{1000} = 0.1\ \text{ms}^{-2}.$ …

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Concept: Differentiate the Cyclist's Position Vector Twice Along the Arc, Then Evaluate at $\theta=45^\circ$

Method: Explicit Rotating-Position-Vector Derivation, Not the "No Tangential + $v^2/r$ " Recipe

Rather than simply stating "constant speed on a circle means only centripetal acceleration, $a=v^2/r$ ," this method builds the cyclist's position on the arc as an explicit function of the angle swept from P, differentiates it twice with respect to time, and reads the direction of acceleration at R directly off the resulting vector formula.

Step 1 -- Set up the position vector along the arc P $\to$ R $\to$ Q

Put the centre O at the origin, with P due east of O (angle $0^\circ$ from the $x$ -axis) and Q due north of O (angle $90^\circ$ ). As the cyclist rides the arc from P through R to Q, their angular position sweeps $\theta$ from $0^\circ$ to $90^\circ$ . With radius $r=1000\ \text{m}$ :

\vec\rho(\theta) = r\cos\theta\,\hat i + r\sin\theta\,\hat j

Since R is described as midway (north-east) between P and Q on the arc, it corresponds to $\theta=45^\circ$ .

Step 2 -- Convert to a function of time using $\theta=\omega t$ (constant $\omega$ , since speed is constant)

On the circular part of the path, constant speed $v=10\ \text{m s}^{-1}$ on radius $r$ means constant angular speed $\omega = v/r$ . So $\theta(t)=\omega t$ , and:

\vec\rho(t) = r\cos(\omega t)\,\hat i + r\sin(\omega t)\,\hat j

Step 3 -- Differentiate once (velocity), confirming the speed matches the given value

\vec v(t) = -r\omega\sin(\omega t)\hat i + r\omega\cos(\omega t)\hat j, \qquad |\vec v| = r\omega = v = 10\ \text{m s}^{-1}\ \checkmark

Step 4 -- Differentiate again (acceleration) -- this is where the direction comes from automatically

\vec a(t) = \frac{d\vec v}{dt} = -r\omega^2\cos(\omega t)\hat i - r\omega^2\sin(\omega t)\hat j = -\omega^2\big[r\cos(\omega t)\hat i + r\sin(\omega t)\hat j\big] = -\omega^2\,\vec\rho(t)

The bracket is exactly $\vec\rho(t)$ from Step 1 -- so the acceleration is automatically a negative multiple of the position vector (measured from the centre O), for every point on the arc, without needing to separately argue "it must point toward the centre."

Step 5 -- Evaluate at R ( $\theta=45^\circ$ ) …

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

Showing the 12 most recent of 18 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. If an object moves with constant speed in a circular path, what can we say about its velocity?$
(a) It is constant
(b) It is changing
(c) It is zero
(d) It is undefined
›Reveal solutionSolution

Uniform circular motion has constant SPEED but continuously changing VELOCITY, because velocity is a vector that depends on direction too.

Velocity has both magnitude (speed) and direction. In circular motion at constant speed, the object's direction of travel is always tangential to the circle, and this tangent direction keeps rotating as the object goes around. So even though |v| stays the same, the vector v (pointing along the tangent) is continuously changing.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. What is the magnitude of centripetal acceleration for an object moving in a circle of radius r with speed v?$
(a) a = v/r
(b) a = r/v
(c) a = v/r^2
(d) a = r^2/v
›Reveal solutionSolution

The true formula for centripetal (radial) acceleration in circular motion is a = v^2 / r. None of the four printed options exactly reproduces this; option (a) is chosen as the intended answer since it is structurally closest (v in the numerator, r in the denominator) and most consistent with a lost superscript.

For an object moving in a circle of radius r with constant speed v, the centripetal acceleration (directed toward the centre) is derived from the rate of change of the velocity vector's direction:

a_c = v^2 / r

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If a particle executes uniform circular motion, then _________.$
(a) its velocity and the acceleration are constant
(b) its velocity and the speed are constant
(c) its speed and the magnitude of acceleration are constant
(d) its acceleration and the speed are constant
›Reveal solutionSolution
Uniform circular motion has constant speed and constant magnitude of centripetal acceleration, but both the velocity and the acceleration continuously change direction, so they are not themselves constant vectors.

In uniform circular motion, a particle moves on a circular path with a constant speed. Let's examine each quantity:

Velocity: This is a vector with both magnitude and direction. Although the magnitude (speed) stays fixed, the direction of velocity is always tangential to the circle and keeps changing as the particle moves around. So velocity is NOT constant.

Speed: This is just the magnitude of velocity, and by definition of 'uniform' circular motion, this stays constant throughout the motion.

…
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CBSE 2025Set ANNUAL1 markMCQ
$Q. What is the magnitude of centripetal acceleration for an object moving in a circle of radius r with speed v?$
(a) a = v/r
(b) a = r/v
(c) a = v^2/r
(d) a = r^2/v
›Reveal solutionSolution

Centripetal acceleration, which keeps an object moving in a circle by constantly redirecting its velocity toward the centre, is given by a = v^2/r.

For an object moving in a circle of radius r with constant speed v, the velocity direction continuously changes even though the speed doesn't. This change in direction produces an acceleration directed toward the centre of the circle, called centripetal …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Match the following - Column A item: Centripetal acceleration. Pick the matching relation from Column B.$
(a) F.V (Force . Velocity)
(b) T is proportional to sqrt(l)
(c) eta is proportional to 1/(dv/dx)
(d) mu_s = tan(theta)
(e) Y is proportional to 1/l
(f) v^2/r
›Reveal solutionSolution

Centripetal acceleration is given by a_c = v^2/r.

For a particle moving with uniform speed v along a circular path of radius r, its velocity direction constantly changes even though its speed does not. This produces an acceleration directed towards the centre of the circle, called centripetal acceleration, with magni …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. A particle is moving with uniform speed of V on a circular path of radius r. What will be the average acceleration in the interval of time in which particle describes half the circle? (A) V^2/r (B) 2V^2/r (C) 2V^2/(πr) (D) V^2/(πr)$
›Reveal solutionSolution

Average acceleration over half a circular revolution is $2V^2/(\pi r)$ .

At the start of the half-circle the velocity is $\vec v_1=V$ in some direction; after half a revolution the particle moves diametrically opposite, so $\vec v_2=-V$ along the original direction. The change in velocity is $|\Delta\vec v|=|{-V}-V|=2V$ .

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. If an object moves with constant speed in a circular path, what can we say about its velocity?$
(a) It is constant
(b) It is changing
(c) It is zero
(d) It is undefined
›Reveal solutionSolution

Velocity = speed + direction. In uniform circular motion, speed is constant but direction changes every instant, so velocity is always changing.

A particle in uniform circular motion has its velocity vector always tangent to the circle. As the particle moves around the circle, the direction of this tangent keeps changing, even though its magnitude (the speed) stays fixed. Since velocity is a vector quantity, a change in direction alone means the v …

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CBSE 2024Set SET-AP55001 markMCQ
$Q. In uniform circular motion:$
(a) Both velocity and acceleration change
(b) Both velocity and acceleration are constant
(c) Velocity remains constant and acceleration changes
(d) Acceleration remains constant and velocity changes
›Reveal solutionSolution

In uniform circular motion, speed and the magnitude of acceleration stay constant, but velocity and acceleration are VECTORS whose direction changes every instant — so both are said to 'change'.

Consider a particle moving on a circle of radius r with constant speed v. At every point, its velocity is tangent to the circle, and as the particle moves around, the direction of this tangent keeps rotating. So even though |v| is fixed, the velocity vector v (which has both magnitude and direction) is continuously changing.

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Velocity vector and acceleration vector of a body in a uniform circular motion are related as$
(1) both in the same direction
(2) perpendicular to each other
(3) both in opposite directions
(4) not related to each other
›Reveal solutionSolution

In uniform circular motion the velocity is tangent to the circle while the acceleration points toward the centre, so they are always perpendicular.

In uniform circular motion, the speed |v| is constant but the direction of v keeps changing, which means there must be an acceleration even though the speed doesn't change. This acceleration is the centripetal acceleration, a_c = v^2/r, and it always points toward the centre of the circle (radially inward).

…

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CBSE 2023Set ANNUAL1 mark
$Q. Answer in one word/sentence: Write the formula for centripetal force for circular motion.$
›Reveal solutionSolution

Centripetal force, the force that keeps a body moving in a circle, is given by F = mv^2/r (equivalently F = momega^2*r).

For an object of mass m moving with constant speed v along a circular path of radius r, its velocity direction is continuously changing, which means it has a centripetal (centre-seeking) acceleration a = v^2/r, directed toward the centre of the circle. By Newton's second law, the net force …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Force acting uniformly on an object doing circular motion is:$
(a) centripetal force
(b) atomic force
(c) internal force
(d) gravitational force
›Reveal solutionSolution

The force in uniform circular motion is the centripetal force.

A body in uniform circular motion has constant speed but continually changing direction, so it is accelerating toward the centre (centripetal acceleration v^2/r). By Newton's second law a …

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CBSE 2022Set TERM11 markMCQ
$Q. A body is travelling in a circle at a constant speed. It$
(1) has an inward acceleration
(2) has constant velocity
(3) has no acceleration
(4) has an outward radial acceleration
›Reveal solutionSolution

Constant speed on a circular path does not mean zero acceleration -- the direction of velocity keeps changing, and this change requires a centripetal (inward) acceleration, v^2/r, at every instant.

Even though the SPEED (magnitude of velocity) is constant, the VELOCITY (a vector) is continuously changing direction as the body goes around the circle. Since acceleration is the rate of change of the velocity VECTOR, a change in direction alone is enough to produce a nonzero acceleration.

…

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