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Class 11 Physics Notes

Motion in a Plane Class 11 Notes

Complete, exam-ready notes on scalars and vectors, vector addition and resolution, motion in a plane with constant acceleration, projectile motion and uniform circular motion — written from the NCERT chapter for CBSE boards, JEE and NEET revision.

Class11SubjectPhysicsCoversCBSE · JEE · NEET

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What is motion in a plane in one line?

Motion in a plane is the study of an object moving in two dimensions (along two perpendicular directions at once) using vectors — here the direction of motion matters, so position, displacement, velocity and acceleration are all treated as vectors. Its two major applications are projectile motion and uniform circular motion.

Scalars and Vectors

Scalar quantity

A quantity that has only a magnitude (a number with a unit) and no direction. Examples: distance, speed, mass, temperature, time, volume and density. Scalars combine by the rules of ordinary algebra.

Vector quantity

A quantity that has both a magnitude and a direction and obeys the triangle law (or equivalently the parallelogram law) of addition. Examples: displacement, velocity, acceleration and force. In handwritten work a vector is marked with an arrow over the letter, e.g. v⃗\vec{v}, and its magnitude is written ∣v⃗∣=v|\vec{v}| = v.

  • Description:\text{Description:}
  • scalar has magnitude only; vector has magnitude and direction.
  • Addition:\text{Addition:}
  • scalar follows ordinary algebra; vector follows triangle / parallelogram laws.
  • Direction:\text{Direction:}
  • scalar needs none; vector needs a specified direction.
  • Examples:\text{Examples:}
  • scalar — speed, mass, temperature; vector — velocity, force, displacement.

The key difference

A direction is associated with a vector but not with a scalar. This is why path length (scalar, always positive) and displacement (vector, can be negative) are different quantities.

MCQ Questions

1Which of the following is a scalar quantity?

2A vector quantity is one that has:

3Ordinary algebraic addition applies to:

4Which of the following is a vector?

5The magnitude of a vector is always:

1 Mark Questions

  1. What is a scalar quantity? Give two examples.
  2. Define a vector quantity and give two examples.
  3. Is distance a scalar or a vector quantity? Justify in one line.
  4. Is the magnitude of a vector always a scalar?
  5. Give one example each of a scalar and a vector that are everyday quantities.

2 Mark Questions

  1. Distinguish between scalar and vector quantities with two examples each.
  2. Can the magnitude of a vector ever be negative? Explain.
  3. Give two physical situations where distance travelled differs from the magnitude of displacement.

3 Mark Questions

  1. Prove that the magnitude of displacement can be zero even when the path length is non-zero, with a suitable example.
  2. Explain with two examples why vectors follow the triangle/parallelogram law of addition while scalars follow ordinary algebra.

Position and Displacement Vectors

Position vector

The vector that locates a particle relative to the origin of a reference frame. For a point P in a plane with coordinates (x, y), the position vector is r⃗=x i^+y j^\vec{r} = x\,\hat{i} + y\,\hat{j}. In three dimensions it extends to r⃗=x i^+y j^+z k^\vec{r} = x\,\hat{i} + y\,\hat{j} + z\,\hat{k}.

Displacement vector

Δr⃗=r⃗ ′−r⃗=(x′−x)i^+(y′−y)j^=Δx i^+Δy j^\Delta \vec{r} = \vec{r}\,' - \vec{r} = (x'-x)\hat{i} + (y'-y)\hat{j} = \Delta x\,\hat{i} + \Delta y\,\hat{j}

The change in position when the particle moves from r⃗\vec{r} to r⃗ ′\vec{r}\,':

Displacement vs path length

Displacement depends only on the end points and can be zero (moving out and back to the start gives a null vector). Path length depends on the actual path and is never zero unless the object is at rest. The two are equal only when motion is along a straight line in one direction.

MCQ Questions

1The position vector of a point (3, 4) is:

2The displacement between P(x₁, y₁) and Q(x₂, y₂) is:

3Displacement is:

1 Mark Questions

  1. Define the position vector of a particle in a plane.
  2. What is meant by the displacement vector of a particle?
  3. Write the displacement vector in component form when a particle moves from (x₁, y₁) to (x₂, y₂).

2 Mark Questions

  1. A particle moves from r = 3î + 2ĵ to r′ = 6î + 6ĵ. Find its displacement vector and its magnitude.
  2. Under what condition is the magnitude of displacement equal to the path length?
  3. Is the displacement of a particle independent of the path taken? Support your answer.

3 Mark Questions

  1. Show that displacement depends only on the end points while path length depends on the actual path, and hence that average speed is always greater than or equal to the magnitude of average velocity.
  2. A particle starting at the origin moves along the path to (3, 4) then to (3, 3). Find its displacement and path length.

Multiplication of a Vector by a Real Number

Multiplying a vector A⃗\vec{A} by a positive real number λ\lambda gives a vector whose magnitude changes by the factor λ\lambda but whose direction is unchanged. Multiplying by a negative number reverses the direction:

∣λA⃗∣=λ∣A⃗∣(λ>0)|\lambda \vec{A}| = \lambda |\vec{A}| \quad(\lambda > 0)
magnitude scales with λ

Physical meaning

Multiplying a velocity vector by a time interval gives a displacement vector — the product of a vector and a scalar that carries its own dimension.

MCQ Questions

1Multiplying a vector by a negative real number:

2Multiplying a vector by zero gives:

3If λ = 3 and A has magnitude 4, then |3A| is:

1 Mark Questions

  1. What is the effect of multiplying a vector by a negative real number?
  2. What vector do you get when a vector is multiplied by zero?
  3. Does multiplying a vector by a positive number change its direction?

2 Mark Questions

  1. A vector A has magnitude 5 units. Find the magnitude and direction of 3A and −2A.
  2. Show that multiplying a velocity vector by an infinitesimal time interval gives a displacement vector.
  3. A force F = 6î N is doubled. Write the new force vector.

3 Mark Questions

  1. Show that multiplying a vector A by a real number λ scales its magnitude by |λ| and reverses its direction when λ < 0.
  2. Two vectors a and b are such that a = λb for a positive λ. What does this tell us about their relative direction and magnitude?

Addition and Subtraction of Vectors

Vectors are added graphically. In the head-to-tail (triangle) method, place the tail of each next vector at the head of the previous one; the resultant R⃗\vec{R} runs from the tail of the first to the head of the last. Equivalently, the parallelogram method draws both vectors from a common origin and takes the diagonal as the sum.

Vector addition is commutative and associative:

A⃗+B⃗=B⃗+A⃗(A⃗+B⃗)+C⃗=A⃗+(B⃗+C⃗)\vec{A} + \vec{B} = \vec{B} + \vec{A}\qquad (\vec{A}+\vec{B})+\vec{C} = \vec{A} + (\vec{B}+\vec{C})

Subtraction is defined as addition of the negative vector: A⃗−B⃗=A⃗+(−B⃗)\vec{A} - \vec{B} = \vec{A} + (-\vec{B}). Adding A⃗\vec{A} and −A⃗-\vec{A} gives the null vector 0⃗\vec{0}, which has zero magnitude and therefore no defined direction.

R2=A2+B2+2ABcos⁡θR^2 = A^2 + B^2 + 2AB\cos\theta
magnitude of the resultant (law of cosines)
tan⁡α=Bsin⁡θA+Bcos⁡θ\tan\alpha = \frac{B\sin\theta}{A + B\cos\theta}
direction of the resultant

Law of sines

The same geometry gives the law of sines Rsin⁡θ=Bsin⁡α=Asin⁡β\frac{R}{\sin\theta} = \frac{B}{\sin\alpha} = \frac{A}{\sin\beta}, which is used to find the direction of the resultant.

MCQ Questions

1Vector addition is:

2The magnitude of R = A + B is maximum when the angle between A and B is:

3The magnitude of R = A + B is minimum when the angle between A and B is:

4A null vector is a vector with:

5For A and B of fixed magnitude, the maximum value of |A + B| is:

1 Mark Questions

  1. State the triangle law of vector addition.
  2. Is vector addition commutative? Write the corresponding relation.
  3. Define a null vector and give one example.
  4. For two vectors of given magnitudes, when is the magnitude of their resultant maximum?

2 Mark Questions

  1. State the parallelogram law of vector addition and write the expression for the magnitude of the resultant.
  2. Two vectors of magnitudes 3 and 4 units act at right angles. Find the magnitude of their resultant.
  3. Define a null vector with two physical examples.
  4. Two vectors A and B of equal magnitude 5 units act at 0°. Find the magnitude of their resultant.

3 Mark Questions

  1. Two vectors A and B of magnitudes 3 and 4 units act at 60°. Find the magnitude and direction of their resultant.
  2. Prove that R² = A² + B² + 2AB cosθ for the resultant of two vectors A and B with an angle θ between them.
  3. Under what condition is |A + B| = |A| + |B|? And |A + B| = ||A| − |B||? Give the angles in each case.

Resolution of Vectors and Unit Vectors

Unit vector

A vector of magnitude one that points in a particular direction; it has no dimension or unit and specifies direction only. The unit vectors along the x-, y- and z-axes are i^, j^, k^\hat{i},\ \hat{j},\ \hat{k}. A general vector can be written as A⃗=∣A⃗∣ n^\vec{A} = |\vec{A}|\,\hat{n} where n^\hat{n} is a unit vector along A⃗\vec{A}.

A vector in a plane can be resolved into rectangular components along the x- and y-axes. If A⃗\vec{A} makes an angle θ\theta with the x-axis, then:

A⃗=Axi^+Ayj^Ax=Acos⁡θ,Ay=Asin⁡θ\vec{A} = A_x\hat{i} + A_y\hat{j}\qquad A_x = A\cos\theta,\quad A_y = A\sin\theta
A=∣A⃗∣=Ax2+Ay2tan⁡θ=AyAxA = |\vec{A}| = \sqrt{A_x^2 + A_y^2}\qquad\tan\theta = \frac{A_y}{A_x}

Components are scalars

Ax and Ay are scalar components; Axi^A_x\hat{i} and Ayj^A_y\hat{j} are the actual component vectors. Magnitude comes from the Pythagorean sum, and the direction from tan⁡θ=Ay/Ax\tan\theta = A_y/A_x.

MCQ Questions

1A unit vector:

2The x-component of a vector A making an angle θ with the x-axis is:

3|î| and |ĵ| are each equal to:

4The unit vector along A = 3î + 4ĵ is:

1 Mark Questions

  1. Define a unit vector. What is the magnitude of î?
  2. Write the expressions for the x- and y-components of a vector A making an angle θ with the x-axis.
  3. Write the formula for the unit vector along a vector A.

2 Mark Questions

  1. A vector of magnitude 10 units makes 60° with the x-axis. Find its x- and y-components.
  2. A = 3î + 4ĵ. Find its magnitude and the angle it makes with the x-axis.
  3. A vector makes 90° with the x-axis. What are its components?

3 Mark Questions

  1. Show that any vector A can be written as A = |A| n̂, where n̂ is the unit vector along A, and hence derive the component resolution of a vector at an angle θ with the x-axis.
  2. A vector of magnitude 20 units has an x-component of 10 units. Find the angle it makes with the x-axis and its y-component.

Vector Addition — Analytical Method

Adding vectors by components is easier and more accurate than drawing them. If A⃗=Axi^+Ayj^\vec{A} = A_x\hat{i}+A_y\hat{j} and B⃗=Bxi^+Byj^\vec{B} = B_x\hat{i}+B_y\hat{j}, then the resultant R⃗=A⃗+B⃗\vec{R} = \vec{A}+\vec{B} has components that are the sums of the corresponding components:

R⃗=Rxi^+Ryj^Rx=Ax+Bx,Ry=Ay+By\vec{R} = R_x\hat{i} + R_y\hat{j}\qquad R_x = A_x + B_x,\quad R_y = A_y + B_y

The same idea extends to any number of vectors. For example, T⃗=a⃗+b⃗−c⃗\vec{T} = \vec{a}+\vec{b}-\vec{c} has components Tx=ax+bx−cxT_x = a_x+b_x-c_x, Ty=ay+by−cyT_y = a_y+b_y-c_y. Once the components of the resultant are known, its magnitude is Rx2+Ry2\sqrt{R_x^2+R_y^2} and its direction is tan⁡θ=Ry/Rx\tan\theta = R_y/R_x.

MCQ Questions

1The analytical method of vector addition uses:

2If R = A + B, then Rₓ equals:

3The magnitude of the resultant R with components Rₓ and Rᵧ is:

4The direction θ of the resultant R is given by:

1 Mark Questions

  1. What do we mean by the analytical method of adding vectors?
  2. Write the component relations for the resultant R = A + B.
  3. Write the formula for the magnitude and direction of the resultant in terms of its components.

2 Mark Questions

  1. A = 2î + 3ĵ and B = 4î − ĵ. Find the resultant R = A + B and its magnitude.
  2. Find the magnitude of the resultant of A = 3î + 4ĵ and B = 4î − 3ĵ.
  3. Why is the analytical method considered more accurate than the graphical method?

3 Mark Questions

  1. Add the three vectors a = 2î + ĵ, b = 3î + 2ĵ and c = î − 4ĵ analytically, and find the magnitude and direction of the resultant.
  2. Explain, step by step, how the analytical (component) method is used to add any number of vectors, and why it is preferred over the graphical method.

Motion in a Plane: Position, Velocity and Acceleration

Position vector: r⃗=xi^+yj^\vec{r} = x\hat{i} + y\hat{j}. The average velocity is the displacement over the time interval, and the (instantaneous) velocity is its limit as Δt→0\Delta t\to 0:

v⃗=lim⁡Δt→0Δr⃗Δt=dr⃗dt\vec{v} = \lim_{\Delta t\to 0}\frac{\Delta \vec{r}}{\Delta t} = \frac{d\vec{r}}{dt}
v⃗=vxi^+vyj^vx=dxdt,vy=dydt\vec{v} = v_x\hat{i} + v_y\hat{j}\qquad v_x = \frac{dx}{dt},\quad v_y = \frac{dy}{dt}
∣v⃗∣=vx2+vy2tan⁡θ=vyvx|\vec{v}| = \sqrt{v_x^2 + v_y^2}\qquad\tan\theta = \frac{v_y}{v_x}

Velocity is always tangent to the path

At any point on the trajectory, the velocity vector is tangential to the path in the direction of motion. This is the property used to find velocity at any instant once the x- and y-coordinates are known as functions of time.

Acceleration is the time rate of change of velocity:

a⃗=lim⁡Δt→0Δv⃗Δt=dv⃗dta⃗=axi^+ayj^\vec{a} = \lim_{\Delta t\to 0}\frac{\Delta \vec{v}}{\Delta t} = \frac{d\vec{v}}{dt}\qquad \vec{a} = a_x\hat{i} + a_y\hat{j}

Velocity and acceleration need not be collinear

In one dimension velocity and acceleration always lie along the same line. In a plane they can have any angle between 0° and 180° — this is what makes projectile and circular motion different from straight-line motion.

MCQ Questions

1The instantaneous velocity of a particle is:

2At any instant the velocity vector is always ___ to the path:

3For motion in a plane, the position vector is written as:

4If r = t²î + 2tĵ, then the velocity at t = 1 s is:

1 Mark Questions

  1. Write the position vector of a particle at (x, y) in a plane.
  2. State the relation that defines instantaneous velocity in terms of the position vector.
  3. Write the instantaneous acceleration in terms of the velocity vector.

2 Mark Questions

  1. The position of a particle is r = 3î + 2tĵ. Find its velocity at any instant.
  2. Show that the velocity vector of a moving particle is always tangential to its path.
  3. For r = 4tî + 3ĵ, find the velocity and state whether the motion is accelerated.

3 Mark Questions

  1. The position of a particle is r(t) = t²î + 2tĵ. Find its velocity and acceleration at t = 2 s.
  2. A particle moves such that r = (t²)î + (2t)ĵ. Find its speed at t = 1 s and its acceleration (which is constant).

Motion in a Plane with Constant Acceleration

When the acceleration a⃗\vec{a} is constant, the vector equations mirror the one-dimensional ones and every component evolves independently:

v⃗=v⃗0+a⃗ tr⃗=r⃗0+v⃗0 t+12a⃗ t2\vec{v} = \vec{v}_0 + \vec{a}\,t\qquad \vec{r} = \vec{r}_0 + \vec{v}_0\,t + \tfrac{1}{2}\vec{a}\,t^2
x=x0+v0xt+12axt2y=y0+v0yt+12ayt2x = x_0 + v_{0x}t + \tfrac{1}{2}a_x t^2\qquad y = y_0 + v_{0y}t + \tfrac{1}{2}a_y t^2

Independent axes

The single most useful result: motion in a plane is a superposition of two independent simultaneous one-dimensional motions along two perpendicular directions (say x and y). Solve each axis with the familiar kinematic equations and combine the results.

MCQ Questions

1For motion with constant acceleration a, the velocity obeys:

2The position for constant acceleration is:

3Under constant acceleration, the x-component of velocity is given by:

1 Mark Questions

  1. Write the vector equation for velocity in motion with constant acceleration.
  2. Write the vector equation for position in motion with constant acceleration.
  3. Write the x-component form of the displacement equation for constant acceleration.

2 Mark Questions

  1. A particle starts from rest with constant acceleration a = (2î + 3ĵ) m/s². Find its velocity after 4 s.
  2. A particle has initial velocity v₀ = 2î m/s and constant acceleration a = î m/s². Write its velocity after time t.
  3. Can the equations of uniform motion loss be applied to projectile motion? Which components remain constant?

3 Mark Questions

  1. Show that in motion with constant acceleration in a plane, the motions along the x- and y-directions can be treated independently.
  2. A particle has initial velocity 2î m/s and constant acceleration (1î + 3ĵ) m/s². Find its velocity and displacement after 3 s.

Projectile Motion

Projectile

An object that is in flight after being thrown or projected (a football, cricket ball, baseball, etc.). Ignoring air resistance, its motion is the combination of a horizontal component with no acceleration and a vertical component with constant acceleration gg downward.

Projected with initial speed v0v_0 at an angle θ0\theta_0 with the horizontal, the components of the initial velocity are v0x=v0cos⁡θ0v_{0x} = v_0\cos\theta_0 and v0y=v0sin⁡θ0v_{0y} = v_0\sin\theta_0. Taking the initial position as origin:

x=(v0cos⁡θ0)ty=(v0sin⁡θ0)t−12gt2x = (v_0\cos\theta_0)t\qquad y = (v_0\sin\theta_0)t - \tfrac{1}{2}gt^2
vx=v0cos⁡θ0vy=v0sin⁡θ0−gtv_x = v_0\cos\theta_0\qquad v_y = v_0\sin\theta_0 - gt

vx is constant

The horizontal velocity component never changes; only vy varies, exactly like an object in free fall. At the top of the path vy = 0.

Eliminating time between x and y gives the equation of the path — a parabola:

y=xtan⁡θ0−gx22v02cos⁡2θ0y = x\tan\theta_0 - \frac{g x^2}{2 v_0^2\cos^2\theta_0}
  • Time to maximum height:  tm=v0sin⁡θ0g\text{Time to maximum height:}\; t_m = \dfrac{v_0\sin\theta_0}{g}
  • Time of flight:  Tf=2v0sin⁡θ0g\text{Time of flight:}\; T_f = \dfrac{2v_0\sin\theta_0}{g}
  • Maximum height:  hm=v02sin⁡2θ02g\text{Maximum height:}\; h_m = \dfrac{v_0^2\sin^2\theta_0}{2g}
  • Horizontal range:  R=v02sin⁡2θ0g\text{Horizontal range:}\; R = \dfrac{v_0^2\sin 2\theta_0}{g}

Range is maximum at 45°

For a fixed launch speed, R is largest when sin⁡2θ0=1\sin 2\theta_0 = 1, i.e. θ0=45∘\theta_0 = 45^\circ, giving Rmax⁡=v02/gR_{\max} = v_0^2/g. Also, angles that exceed or fall short of 45° by the same amount give equal ranges (Galileo's result).

MCQ Questions

1The time of flight of a projectile is:

2The maximum range of a projectile occurs at an angle of:

3The trajectory of a projectile is:

4The maximum height of a projectile is:

5The horizontal range of a projectile is:

1 Mark Questions

  1. Define a projectile.
  2. Write the expression for the time of flight of a projectile.
  3. Write the expression for the maximum height reached by a projectile.
  4. Write the expression for the horizontal range of a projectile.

2 Mark Questions

  1. A projectile is launched with speed v₀ at 45°. Write the expression for its horizontal range.
  2. A ball is thrown with a speed of 20 m/s at 30° to the horizontal. Find its time of flight (g = 10 m/s²).
  3. State the condition for maximum range and give the value of the range in that case.
  4. At what angle should a projectile be fired so that its horizontal range equals the maximum possible value?

3 Mark Questions

  1. A ball is thrown at 20 m/s at 30° above the horizontal. Find its time of flight, maximum height and horizontal range (g = 10 m/s²).
  2. Show that for a given launch speed the range is maximum at 45°, and that angles symmetric about 45° give the same range.
  3. Show that the trajectory of a projectile is a parabola, starting from y = x tanθ₀ − gx²/(2v₀²cos²θ₀).

Uniform Circular Motion

Uniform circular motion

Motion of an object along a circular path at constant speed. The word 'uniform' refers to the speed, which is constant — the velocity is still changing because its direction changes continuously, so the object is accelerating.

The velocity is always tangential to the circle, and the acceleration is directed towards the centre. This centre-seeking acceleration is the centripetal acceleration:

ac=v2Ra_c = \frac{v^2}{R}

Angular speed

ω=ΔθΔt\omega = \frac{\Delta\theta}{\Delta t}

The time rate of change of angular displacement. It is linked to linear speed and to centripetal acceleration by:

v=ωRac=ω2Rv = \omega R\qquad a_c = \omega^2 R

The time for one revolution is the time period T, and the frequency (revolutions per second) is ν=1/T\nu = 1/T:

v=2πRT=2πRνω=2πνac=4π2ν2Rv = \frac{2\pi R}{T} = 2\pi R\nu\qquad \omega = 2\pi\nu\qquad a_c = 4\pi^2\nu^2 R

Not a constant vector

Although the magnitude of ac=v2/Ra_c = v^2/R is constant, its direction always points to the centre and changes continuously. So centripetal acceleration is not a constant vector, and the constant-acceleration kinematic equations do NOT apply to uniform circular motion.

MCQ Questions

1The centripetal acceleration is:

2The linear speed is related to angular speed by:

3The direction of centripetal acceleration is:

4For uniform circular motion, the angular speed is:

5The angular speed for time period T is:

1 Mark Questions

  1. Define uniform circular motion.
  2. Write the expression for centripetal acceleration in terms of speed and radius.
  3. Define angular speed ω and give its unit.
  4. Write the relation between angular speed, linear speed and radius.

2 Mark Questions

  1. A body moves in a circle of radius 2 m with a speed of 4 m/s. Find its centripetal acceleration.
  2. State why the velocity of a body in uniform circular motion is not constant even though its speed is constant.
  3. A particle moves in a circle of radius 1 m with angular speed 2 rad/s. Find its linear speed.
  4. Express angular speed in terms of frequency ν and time period T.

3 Mark Questions

  1. Show that centripetal acceleration is aᴄ = v²/R = ω²R, and explain why its direction is always towards the centre.
  2. A particle makes 10 revolutions in 20 s in a circle of radius 10 cm. Find its angular speed, linear speed and centripetal acceleration.
  3. Two particles move in circles of radii R and 2R with the same angular speed. Compare their linear speeds and centripetal accelerations.

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Vector resolution

Ax=Acos⁡θ,Ay=Asin⁡θA_x = A\cos\theta,\quad A_y = A\sin\theta

|A| = √(Ax² + Ay²)

Resultant magnitude

R2=A2+B2+2ABcos⁡θR^2 = A^2 + B^2 + 2AB\cos\theta

Law of cosines

Analytical addition

Rx=Ax+Bx,Ry=Ay+ByR_x = A_x + B_x,\quad R_y = A_y + B_y

Add corresponding components

Position vector

r⃗=xi^+yj^\vec{r} = x\hat{i} + y\hat{j}

Displacement Δr = r′ − r

Velocity

v⃗=dr⃗dt\vec{v} = \frac{d\vec{r}}{dt}

Components vx = dx/dt, vy = dy/dt

Constant acceleration

v⃗=v⃗0+a⃗t,r⃗=r⃗0+v⃗0t+12a⃗t2\vec{v} = \vec{v}_0 + \vec{a}t,\quad \vec{r} = \vec{r}_0 + \vec{v}_0t + \tfrac12 \vec{a}t^2

Axes independent

Projectile trajectory

y=xtan⁡θ0−gx22v02cos⁡2θ0y = x\tan\theta_0 - \frac{gx^2}{2v_0^2\cos^2\theta_0}

A parabola

Projectile range / height

R=v02sin⁡2θ0g,hm=v02sin⁡2θ02gR = \frac{v_0^2\sin 2\theta_0}{g},\quad h_m = \frac{v_0^2\sin^2\theta_0}{2g}

Maximum range at 45°: R = v₀²/g

Time of flight

Tf=2v0sin⁡θ0gT_f = \frac{2v_0\sin\theta_0}{g}

Tf = 2tm

Centripetal acceleration

ac=v2R=ω2Ra_c = \frac{v^2}{R} = \omega^2 R

Towards the centre

Angular ↔ linear speed

v=ωR=2πRT=2πRνv = \omega R = \frac{2\pi R}{T} = 2\pi R\nu

ω = 2πν

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

  • CBSE Board: definitions of scalar/vector, triangle and parallelogram laws, and a short numerical on projectile range or centripetal acceleration are asked year after year.
  • JEE Main: 1–2 questions typically on vector resolution, the resultant of two vectors, projectile motion (range, time of flight) or uniform circular motion.
  • NEET: short numericals on range, maximum height, time of flight and centripetal acceleration — memorise the projectile table and v²/R.
  • Common traps: do not apply constant-acceleration equations to uniform circular motion; remember centripetal acceleration is always directed to the centre; distinguish path length (scalar) from displacement (vector).

Solved problems

Worked examples

JEE / NEET-style numericals, solved step by step.

1

Resultant of a falling rain and a wind

Rain is falling vertically with a speed of 35 m/s. A wind starts blowing from east to west with a speed of 12 m/s. In what direction should a boy at a bus stop hold his umbrella?

  1. The rain and wind velocities are perpendicular, so the resultant has magnitude R = √(vr² + vw²) = √(35² + 12²).
  2. R = √(1225 + 144) = √1369 = 37 m/s.
  3. The angle with the vertical is tan θ = vw/vr = 12/35 ≈ 0.343, so θ ≈ 19°.

Answer

θ≈19∘ with the vertical, towards the east\theta \approx 19^\circ\ \text{with the vertical, towards the east}
2

Velocity and acceleration from the position vector

The position of a particle is r = 3.0t î − 2.0t² ĵ + 4.0 k̂ (metres). Find v(t) and a(t), and the magnitude and direction of v at t = 1.0 s.

  1. Differentiate: v = dr/dt = 3.0 î + (−4.0t) ĵ.
  2. Acceleration: a = dv/dt = −4.0 ĵ → |a| = 4.0 m/s² along the −y direction.
  3. At t = 1.0 s, v = 3.0 î − 4.0 ĵ → |v| = √(3² + 4²) = 5.0 m/s.
  4. Direction: tan θ = vy/vx = −4/3, so θ ≈ −53° with the x-axis.

Answer

∣v⃗∣=5.0 m/s,θ≈−53∘|\vec{v}| = 5.0\,\text{m/s},\quad \theta \approx -53^\circ
3

Cricket ball thrown at 30°

A cricket ball is thrown with a speed of 28 m/s in a direction 30° above the horizontal. Find (a) the maximum height, (b) the time to return to the same level, and (c) the range.

  1. (a) h = v₀² sin²θ / 2g = (28² × sin²30°) / (2 × 9.8) = (784 × 0.25)/19.6 = 10.0 m.
  2. (b) T = 2v₀ sinθ / g = (2 × 28 × 0.5)/9.8 = 28/9.8 ≈ 2.9 s.
  3. (c) R = v₀² sin 2θ / g = (28² × sin 60°)/9.8 = (784 × 0.866)/9.8 ≈ 69 m.

Answer

hm=10.0 m,Tf≈2.9 s,R≈69 mh_m = 10.0\,\text{m},\quad T_f \approx 2.9\,\text{s},\quad R \approx 69\,\text{m}
4

Insect in a circular groove (uniform circular motion)

An insect trapped in a circular groove of radius 12 cm makes 7 revolutions in 100 s. Find (a) its angular speed and linear speed, and (b) the magnitude of its acceleration.

  1. Frequency ν = 7/100 = 0.07 rev/s, so angular speed ω = 2πν ≈ 0.44 rad/s.
  2. Linear speed v = ωR = 0.44 × 12 = 5.3 cm/s, tangential to the circle.
  3. Centripetal acceleration a = ω²R = (0.44)² × 12 ≈ 2.3 cm/s², directed towards the centre.

Answer

ω≈0.44 rad/s,v≈5.3 cm/s,a≈2.3 cm/s2\omega \approx 0.44\,\text{rad/s},\quad v \approx 5.3\,\text{cm/s},\quad a \approx 2.3\,\text{cm/s}^2

FAQ

Common questions

What is the difference between a scalar and a vector?

A scalar has only magnitude (e.g. speed, mass, temperature) and follows ordinary algebra. A vector has both magnitude and direction (e.g. velocity, force, displacement) and follows the triangle or parallelogram law of addition.

Why is the path of a projectile a parabola?

After eliminating time between x = v0cosθ0·t and y = v0sinθ0·t − ½gt², you get y = x tanθ0 − gx²/(2v0²cos²θ0), which has the form y = ax + bx² — the equation of a parabola.

At what angle is the range of a projectile maximum, and why?

The range is R = v0² sin2θ0/g. For a fixed launch speed it is maximum when sin2θ0 = 1, i.e. θ0 = 45°, giving Rmax = v0²/g. Angles equispaced around 45° give equal ranges.

Why does an object moving at constant speed in a circle accelerate?

Acceleration is the rate of change of velocity, and velocity has both magnitude and direction. Even though the speed (magnitude) is constant, the direction is always changing, so the velocity is changing and hence there is a centripetal acceleration v²/R directed towards the centre.

Can the constant-acceleration equations be used for uniform circular motion?

No. In uniform circular motion the magnitude of acceleration is constant but its direction changes continuously, so the acceleration is not constant and the kinematic equations v = v0 + at and r = r0 + v0t + ½at² do not apply.

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