Kinematics: Equations of Motion, Projectile, Relative Velocity

Everything around us is in motion – a ball rolling on the ground, a train speeding between stations, a cricket ball flying towards the boundary, even the Earth spinning on its axis. The branch of physics that describes motion – how far, how fast and in which direction an object moves – without asking what causes the motion is called kinematics. The word comes from the Greek kinema, meaning motion.

Kinematics is a regular source of questions in SSC CGL, CHSL, MTS, GD, RRB NTPC, Group D, ALP, JE, WBCS, Police and State PSC exams. Questions are asked on distance and displacement, speed and velocity, acceleration, the three equations of motion, graphs, free fall, projectile motion, relative velocity and circular motion – often as short numerical problems. This guide explains each idea in simple language with diagrams, formulas, solved examples, memory tricks and 20 MCQs with answers.

➤ Scalars vs vectors: Distance and speed are scalars · Displacement, velocity and acceleration are vectors
➤ Equations of motion: v = u + at · s = ut + ½at² · v² = u² + 2as
➤ Graphs: Slope of x–t graph = velocity · Slope of v–t graph = acceleration · Area under v–t graph = displacement
➤ Free fall: Acceleration due to gravity g ≈ 9.8 m/s² (often taken as 10 m/s²), same for all bodies in vacuum
➤ Projectile: Range R = u² sin 2θ / g · Maximum range at 45° · Complementary angles give the same range
➤ Relative velocity: Same direction → subtract speeds · Opposite direction → add speeds
➤ Uniform circular motion: Speed constant but velocity changes · Centripetal acceleration a = v²/r

Basic Terms of Motion

An object is said to be in motion if its position changes with time compared with a reference point, and at rest if it does not. Rest and motion are relative – a passenger sitting in a moving train is at rest relative to the train but in motion relative to the platform.

Key points: Distance can never be less than displacement. Displacement can be zero even when distance is not – for example, after one complete round of a circular track. Negative acceleration (slowing down) is called retardation or deceleration. To convert km/h into m/s, multiply by 5/18; to convert m/s into km/h, multiply by 18/5.

Example: A runner completes one full round of a circular track of radius 7 m. Distance = 2πr = 2 × 22/7 × 7 = 44 m; displacement = 0, because she returns to the starting point.

Types of Motion

Motion in One Dimension and the Equations of Motion

When an object moves along a straight line with constant (uniform) acceleration, its motion can be described by three simple equations. Here u = initial velocity, v = final velocity, a = acceleration, t = time and s = displacement.

Example 1: A car starts from rest and accelerates uniformly at 2 m/s² for 10 seconds. Find its final velocity and the distance covered.
Here u = 0, a = 2 m/s², t = 10 s.
v = u + at = 0 + 2 × 10 = 20 m/s (= 72 km/h)
s = ut + ½at² = 0 + ½ × 2 × 10² = 100 m

Example 2: A train moving at 72 km/h applies brakes and stops in 100 m. Find the retardation.
u = 72 × 5/18 = 20 m/s, v = 0, s = 100 m. Using v² = u² + 2as: 0 = 400 + 2 × a × 100, so a = −2 m/s² (retardation of 2 m/s²).

Motion Graphs

Graphs make motion easy to understand. On a position–time (x–t) graph, the slope gives velocity – a straight sloping line means uniform velocity and a horizontal line means the object is at rest. On a velocity–time (v–t) graph, the slope gives acceleration and the area under the graph gives displacement (Figure 1).

Figure 1: Velocity–time graph – the slope gives acceleration and the area under the line gives displacement

Free Fall and Motion Under Gravity

When an object falls only under the pull of gravity, with no air resistance, it is in free fall. All objects – heavy or light – fall with the same acceleration, g ≈ 9.8 m/s², near Earth's surface. Galileo Galilei first showed this. In a vacuum, a feather and a coin fall together. In air, the feather falls slowly only because of air resistance. The value of g is slightly more at the poles than at the equator, and decreases with height above the Earth.

For free fall, use the equations of motion with a = g (downward motion) or a = −g (upward motion):

Example 3: A stone is dropped from a height of 80 m. Taking g = 10 m/s², find the time taken and the velocity on hitting the ground.
t = √(2h/g) = √(160/10) = √16 = 4 s; v = gt = 10 × 4 = 40 m/s.

Motion in Two Dimensions

When an object moves in a plane, such as a ball thrown at an angle, its motion has two parts – horizontal (x) and vertical (y). Velocity, displacement and acceleration then have to be treated as vectors.

Scalars, Vectors and Vector Addition

Vectors are added using the triangle law or the parallelogram law. For two vectors A and B at an angle θ, the resultant is R = √(A² + B² + 2AB cos θ).

• If A and B are in the same direction (θ = 0°): R = A + B (maximum)
• If A and B are opposite (θ = 180°): R = A − B (minimum)
• If A and B are perpendicular (θ = 90°): R = √(A² + B²)

A vector A at angle θ with the x-axis has components A_x = A cos θ and A_y = A sin θ – this is called resolution of a vector.

Projectile Motion

An object thrown into the air at an angle and moving under gravity alone is called a projectile – for example, a cricket ball hit for a six, a javelin, a football kick or water from a hose. Its path is a parabola (Figure 2). The secret of projectile motion is to treat the two directions separately:

• Horizontal motion: No force acts horizontally (ignoring air resistance), so the horizontal velocity u cos θ stays constant.
• Vertical motion: Gravity acts downward, so the vertical velocity u sin θ decreases, becomes zero at the top, and then increases downward – exactly like a ball thrown straight up.

Figure 2: Path of a projectile launched with speed u at angle θ – horizontal velocity stays constant while vertical velocity changes due to gravity

Projectile Formulas

Example 4: A ball is kicked at 20 m/s at an angle of 45°. Taking g = 10 m/s², find the range, maximum height and time of flight.
R = u² sin 2θ / g = 400 × sin 90° / 10 = 40 m
H = u² sin² θ / 2g = 400 × (1/2) / 20 = 10 m (note R = 4H at 45°)
T = 2u sin θ / g = 2 × 20 × 0.707 / 10 ≈ 2.83 s

Horizontal projectile: If a ball is thrown horizontally from a height h (like a ball rolling off a table, or a packet dropped from a flying aircraft), the time to reach the ground is t = √(2h/g), the same as an object simply dropped from that height. The horizontal distance covered is u × √(2h/g). That is why a bullet fired horizontally and a bullet dropped from the same height reach the ground at the same time.

Relative Velocity

The velocity of one object as seen from another moving object is called its relative velocity. The relative velocity of A with respect to B is v_AB = v_A − v_B (as vectors).

Example 5 (train problems): A 200 m long train moving at 72 km/h crosses a pole. Time = length ÷ speed = 200 ÷ 20 = 10 s. If it crosses a 300 m long platform, it must cover 200 + 300 = 500 m, so time = 500 ÷ 20 = 25 s.

River–Boat Problems

A boat crossing a flowing river has two velocities – its own velocity in still water (v_b) and the velocity of the river current (v_r). The boat's actual motion is the vector sum of the two (Figure 3).

Figure 3: A boat steered straight across a river drifts downstream because of the current; its actual path is the resultant of the two velocities

Example 6: A river is 100 m wide and flows at 3 m/s. A boat moves at 5 m/s in still water. (a) If it heads straight across: time = 100/5 = 20 s, drift = 3 × 20 = 60 m. (b) To reach the point directly opposite: sin θ = 3/5, so θ ≈ 37° upstream; speed across = √(25 − 9) = 4 m/s, time = 100/4 = 25 s.

Uniform Circular Motion

When an object moves in a circle at constant speed, its direction keeps changing, so its velocity changes and it is accelerating – even though the speed is constant. This acceleration always points towards the centre and is called centripetal acceleration.

Exam point: In uniform circular motion, speed and kinetic energy remain constant, but velocity and momentum change continuously because of the change in direction. The velocity at any point is along the tangent to the circle – that is why mud flies off a spinning wheel tangentially.

Kinematics Formula Sheet

West Bengal Connection – Motion Around Us

Kolkata gives many everyday examples of motion. The Kolkata Metro, India's first metro railway, began service on 24 October 1984 between Esplanade and Netaji Bhavan (Bhowanipur). In March 2024, the city opened India's first underwater metro section under the Hooghly river between Howrah Maidan and Esplanade on the East–West corridor. A metro train speeding up from a station, running at steady speed and braking before the next stop is a perfect real-life example of acceleration, uniform motion and retardation. The trams of Kolkata, one of the oldest tram systems in Asia, and the ferries crossing the Hooghly are similar everyday examples of uniform motion and relative velocity – a ferry heading straight across the river drifts downstream with the current, exactly as in the river–boat problem above.

For more physics, read our guide on Light Reflection and Refraction, and explore space science in Darbhanga Planetarium's ISRO Space Gallery. Motion of air and wind is covered in Climatology and Atmospheric Circulation. Practise numerical questions with the CSIR NAL MTS Mock Test.

20 MCQs on Kinematics

Q1. Which of the following is a vector quantity?
(a) Speed
(b) Displacement
(c) Distance
(d) Time

Answer: (b) – Displacement has both magnitude and direction; speed, distance and time are scalars.

Q2. A person walks once around a circular park of radius r and returns to the starting point. His displacement is:
(a) Zero
(b) 2πr
(c) 2r
(d) πr

Answer: (a) – He returns to the starting point; the distance covered is 2πr.

Q3. 72 km/h is equal to:
(a) 10 m/s
(b) 15 m/s
(c) 20 m/s
(d) 25 m/s

Answer: (c) – 72 × 5/18 = 20 m/s.

Q4. The slope of a velocity–time graph gives:
(a) Displacement
(b) Speed
(c) Distance
(d) Acceleration

Answer: (d) – Slope of v–t graph = acceleration; the area under it gives displacement.

Q5. The area under a velocity–time graph represents:
(a) Acceleration
(b) Displacement
(c) Force
(d) Momentum

Answer: (b) – For uniform acceleration, the area is a trapezium equal to ut + ½at².

Q6. A car starts from rest and accelerates at 2 m/s². What distance will it cover in 5 seconds?
(a) 10 m
(b) 20 m
(c) 50 m
(d) 25 m

Answer: (d) – s = ut + ½at² = 0 + ½ × 2 × 25 = 25 m.

Q7. Which equation of motion does not contain time (t)?
(a) v² = u² + 2as
(b) v = u + at
(c) s = ut + ½at²
(d) a = (v − u)/t

Answer: (a) – The third equation links velocity and displacement directly.

Q8. In the absence of air resistance, a feather and a coin dropped together from the same height will:
(a) Feather reaches first
(b) Reach the ground together
(c) Coin reaches first
(d) Feather will not fall

Answer: (b) – In vacuum all bodies fall with the same acceleration g, as Galileo showed.

Q9. A stone is dropped from a height of 45 m. How long will it take to reach the ground? (g = 10 m/s²)
(a) 2 s
(b) 4.5 s
(c) 3 s
(d) 9 s

Answer: (c) – t = √(2h/g) = √(90/10) = √9 = 3 s.

Q10. When a ball is thrown vertically upward, at the highest point its:
(a) Velocity and acceleration are both zero
(b) Velocity is maximum
(c) Acceleration is zero but velocity is not
(d) Velocity is zero but acceleration is g

Answer: (d) – Gravity keeps acting even at the top, so acceleration remains g downward.

Q11. The path followed by a projectile thrown at an angle is a:
(a) Straight line
(b) Parabola
(c) Circle
(d) Hyperbola

Answer: (b) – Constant horizontal velocity combined with uniformly accelerated vertical motion gives a parabola.

Q12. A projectile has maximum horizontal range when it is thrown at an angle of:
(a) 30°
(b) 60°
(c) 90°
(d) 45°

Answer: (d) – R = u² sin 2θ / g is maximum when sin 2θ = 1, i.e., θ = 45°.

Q13. Which pair of projection angles gives the same horizontal range (same speed)?
(a) 30° and 45°
(b) 20° and 80°
(c) 30° and 60°
(d) 45° and 90°

Answer: (c) – Complementary angles (adding up to 90°) give equal ranges.

Q14. At the highest point of its path, the velocity of a projectile is:
(a) Zero
(b) u sin θ
(c) u cos θ
(d) u

Answer: (c) – Only the horizontal component remains; the vertical component becomes zero.

Q15. A ball is thrown at 30 m/s at 45°. Its horizontal range is (g = 10 m/s²):
(a) 90 m
(b) 45 m
(c) 60 m
(d) 180 m

Answer: (a) – R = u²/g at 45° = 900/10 = 90 m.

Q16. Two trains move in opposite directions at 50 km/h and 40 km/h. Their relative speed is:
(a) 10 km/h
(b) 45 km/h
(c) 20 km/h
(d) 90 km/h

Answer: (d) – For opposite directions, relative speed = 50 + 40 = 90 km/h.

Q17. A 150 m long train moving at 54 km/h crosses an electric pole in:
(a) 10 s
(b) 15 s
(c) 5 s
(d) 20 s

Answer: (a) – 54 km/h = 15 m/s; time = 150 ÷ 15 = 10 s.

Q18. To cross a river in the shortest possible time, a swimmer should swim:
(a) Upstream at 45°
(b) Perpendicular to the river current
(c) Along the current
(d) Downstream at 45°

Answer: (b) – Swimming straight across uses the full speed to cross; the current only causes drift.

Q19. In uniform circular motion, which quantity remains constant?
(a) Velocity
(b) Acceleration (as a vector)
(c) Speed
(d) Momentum

Answer: (c) – Speed is constant, but velocity, momentum and the direction of acceleration keep changing.

Q20. India's first underwater metro section, opened in 2024, runs beneath which river?
(a) Hooghly
(b) Yamuna
(c) Mithi
(d) Sabarmati

Answer: (a) – The Kolkata Metro East–West corridor runs under the Hooghly between Howrah Maidan and Esplanade.

Want more practice? Try our Daily GK Quiz and Mock Tests.

Conclusion

Kinematics gives us the language to describe motion – distance and displacement, speed and velocity, and acceleration. With just three equations of motion and a clear understanding of graphs, you can solve most straight-line problems, including free fall. Projectile motion becomes easy when horizontal and vertical motions are treated separately, and relative velocity explains everything from overtaking trains to boats crossing a river. For exams, remember the formulas, the 45° rule, complementary angles, unit conversions and the train and river shortcuts. Revise the tricks above and practise the numericals regularly.

Reference: NCERT Science Class IX ("Motion"); NCERT Physics Class XI ("Motion in a Straight Line" and "Motion in a Plane").