Why this topic matters · 8 min read
Kinematics is the foundation of Agniveer Vayu physics—expect 2-3 questions mixing 1D (straight-line) and 2D (projectile) motion. Exams test your ability to apply equations of motion, resolve vectors, and analyze projectile trajectories under gravity. High weightage in numerical problem-solving; focus on relative motion and projectile range/height calculations.
Equations of Motion (1D)
These three equations describe how objects move in a straight line under constant acceleration. They link displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t). Think of them as the 'holy trinity' of kinematics—memorize all three forms because different problems give you different known quantities. For example, if you don't know time, use the equation without t. If you don't know final velocity, use the one without v.
- v = u + at (velocity changes linearly with time)
- s = ut + (1/2)at^2 (displacement includes both initial motion and acceleration effect)
- v^2 = u^2 + 2as (relates velocity and displacement without time)
- Average velocity = (u + v)/2 when acceleration is constant
- Displacement is area under velocity-time graph
- Use sign convention: take one direction as positive, opposite as negative
Key formulas
First equation of motion
v = u + at
When: When time is given or asked; simplest form
Second equation of motion
s = ut + (1/2)at^2
When: When you need displacement and time is known; doesn't involve final velocity
Third equation of motion
v^2 = u^2 + 2as
When: When time is not given or not needed; links velocity and displacement directly
Worked examples
A car starts from rest (u=0) and accelerates at 2 m/s^2 for 5 seconds. Find distance: s = 0 + (1/2)(2)(5^2) = 25 m. Find final velocity: v = 0 + 2(5) = 10 m/s.
A ball is thrown upward with u=20 m/s. Find maximum height (v=0 at top): 0 = 400 + 2(-10)s, so s = 20 m. Time to reach top: 0 = 20 - 10t, so t = 2 s.
Projectile Motion (2D)
Projectile motion is 1D kinematics applied independently to two perpendicular directions: horizontal (x) and vertical (y). The key insight is that horizontal and vertical motions are completely independent—there is no horizontal acceleration (ignoring air resistance), so horizontal velocity stays constant. Vertical motion is just free fall under gravity. Decompose the initial velocity into components using trigonometry, then treat each direction separately.
- Horizontal: x = u_x * t (constant velocity, no acceleration)
- Vertical: y = u_y * t - (1/2)gt^2 (free fall under gravity g = 10 m/s^2)
- Initial velocity components: u_x = u*cos(theta), u_y = u*sin(theta)
- At maximum height: v_y = 0 (vertical velocity becomes zero momentarily)
- Time of flight: total time in air = 2*u_y/g = 2*u*sin(theta)/g
- Range (horizontal distance): R = u^2*sin(2*theta)/g (maximum at 45 degrees)
Key formulas
Horizontal displacement
x = u*cos(theta)*t
When: Finding how far the projectile travels horizontally
Vertical displacement
y = u*sin(theta)*t - (1/2)gt^2
When: Finding height at any time t during flight
Time of flight
T = 2*u*sin(theta)/g
When: Finding total time projectile stays in air (lands at same height as launch)
Range formula
R = u^2*sin(2*theta)/g
When: Finding horizontal distance when projectile lands at launch height
Maximum height
H = u^2*sin^2(theta)/(2g)
When: Finding peak height above launch point
Worked examples
A ball is thrown at 20 m/s at 30 degrees. Range: R = 400*sin(60)/10 = 400*(0.866)/10 = 34.6 m. Time of flight: T = 2*20*sin(30)/10 = 2*20*0.5/10 = 2 seconds.
A projectile is fired horizontally from a cliff at 15 m/s. After 2 seconds: horizontal distance x = 15*2 = 30 m, vertical distance y = 0 - (1/2)*10*4 = -20 m (20 m below launch point).
Relative Motion
Relative motion asks: how does object A move as seen from object B's perspective? The trick is vector subtraction. Velocity of A relative to B is v_AB = v_A - v_B (as vectors). This is crucial for river-crossing problems, rain problems, and aircraft problems in Agniveer exams. Always draw a vector diagram—it clarifies the geometry instantly. Remember: relative velocity is the difference of velocities, not the difference of speeds.
- Relative velocity: v_AB = v_A - v_B (vector subtraction)
- In river problems: boat velocity relative to ground = boat velocity relative to water + water velocity relative to ground
- For perpendicular motion: use Pythagoras to find resultant
- Relative displacement: s_AB = s_A - s_B
- Time to collision/meeting: use relative velocity and relative displacement
- Rain appears to come at an angle due to observer's motion—use velocity triangle
Key formulas
Relative velocity vector
v_AB = v_A - v_B
When: Finding how fast A moves as seen from B
Resultant of perpendicular velocities
v_resultant = sqrt(v_1^2 + v_2^2)
When: When two velocity components are perpendicular
Worked examples
A boat wants to cross a river of width 100 m. Boat speed in water is 10 m/s, river current is 5 m/s. To cross perpendicular to banks: boat must aim at angle such that resultant points straight across. Relative velocity = sqrt(10^2 - 5^2) = sqrt(75) = 8.66 m/s across. Time = 100/8.66 = 11.5 seconds.
Rain falls vertically at 10 m/s. Observer moves horizontally at 6 m/s. Rain appears to come at angle: tan(angle) = 6/10, so angle = 31 degrees from vertical.
Graphs in Kinematics
Velocity-time (v-t) and displacement-time (s-t) graphs are visual shortcuts in Agniveer exams. A v-t graph's slope is acceleration. The area under a v-t graph is displacement. An s-t graph's slope is velocity. A curved s-t graph means acceleration is present. These graphs often appear as 'read the graph and answer' questions—practice interpreting them quickly.
- Slope of s-t graph = velocity
- Slope of v-t graph = acceleration
- Area under v-t graph = displacement
- Straight line in s-t graph means constant velocity
- Curved s-t graph (parabola) means constant acceleration
- Horizontal line in v-t graph means zero acceleration (uniform motion)
⚠ Common mistakes to avoid
- Forgetting to resolve velocity into components in 2D problems—always split u into u*cos(theta) and u*sin(theta) at the start
- Using g = 9.8 m/s^2 when exam expects g = 10 m/s^2—check the question; most Agniveer papers use 10 for simplicity
- Mixing up 'range' with 'maximum height'—range is horizontal distance, height is vertical; they use different formulas
- In relative motion, adding velocities instead of subtracting—v_AB = v_A - v_B, not v_A + v_B
- Ignoring sign convention—treating upward and downward as same sign leads to wrong answers in vertical motion
🧠 Memory aids
- SUVAT: S=displacement, U=initial velocity, V=final velocity, A=acceleration, T=time. Pick the three you know, find the two you need.
- For projectile: 'Horizontal is lazy, vertical is crazy'—horizontal velocity never changes, vertical velocity changes every instant due to gravity.
- Range formula sin(2*theta): Remember 2*theta because sin(2*theta) = 2*sin(theta)*cos(theta), which is why 45 degrees gives maximum range (sin(90) = 1).
- Relative velocity: 'Subtract to see'—v_AB = v_A minus v_B. Think of it as 'what A looks like from B's frame'.
🎯 AGNIVEER VAYU exam tips
- Agniveer Vayu typically gives 1-2 projectile motion questions per paper—focus on range, time of flight, and maximum height calculations. Recent papers favor 30-45-60 degree angles.
- Relative motion appears in 'boat crossing river' or 'rain angle' format—always draw a velocity triangle; it's faster than algebra.
- Graph-based questions are common—practice reading v-t and s-t graphs in 30 seconds. Examiners test if you can extract slope and area quickly.
- Numerical answers are usually 'nice' numbers (10, 20, 25, 50 m, etc.)—if your answer is messy, recheck your calculation or g value.
- Time management: kinematics questions are solvable in 2-3 minutes if you know which equation to use. Spend 10 seconds choosing the right formula, then calculate.
Q1 · medium · AI-verified
A stone is dropped from a height of 80 m. How long does it take to reach the ground? (g = 10 m/s²)
- 4 s
- 2 s
- 8 s
- 6 s
Q2 · hard · AI-verified
A body is thrown vertically upward from the ground. It takes 3 s to reach the highest point. The velocity with which it was thrown and the total distance covered on its return to the ground are respectively (g = 10 m/s²):
- 30 m/s and 90 m
- 30 m/s and 60 m
- 15 m/s and 90 m
- 30 m/s and 45 m
Q3 · hard · AI-verified
A particle starts from rest and moves along a straight line with acceleration a = (4 − 2t) m/s². The particle comes to rest again after how many seconds?
- 6 s
- 2 s
- 4 s
- 8 s
Q4 · hard · AI-verified
Two stones are thrown vertically upward simultaneously from the same point with speeds 20 m/s and 30 m/s. The ratio of the maximum heights reached by them is (g = 10 m/s²):
- 16 : 81
- 1 : 2
- 2 : 3
- 4 : 9
Q5 · medium · AI-verified
A car travels the first half of a journey at 30 km/h and the second half at 60 km/h. What is the average speed for the entire journey?
- 50 km/h
- 45 km/h
- 35 km/h
- 40 km/h