Kinematics explains the motion of objects using displacement, velocity, acceleration, graphs, and equations of motion. For NEET 2027, important topics include projectile motion, relative velocity, river crossing, motion under gravity, and stopping distance. A strong grasp of graph slopes, areas, and sign conventions helps solve questions quickly.
This Kinematics revision covers core concepts, shortcuts, and PYQ-based applications. It explains how differentiation and integration connect position, velocity, and acceleration, and how to approach both constant and variable acceleration problems with confidence. Use the PW’s NEET resources, including PW’s NEET notes and PYQs, concepts, and shortcuts below to build accuracy and speed for the exam.
Distance is the actual length of the path travelled by a particle. It is a scalar quantity.
Displacement is the change in position:
Displacement = Final position - Initial position
Its magnitude is the shortest distance between the initial and final points. Displacement has direction and is treated as a vector.
The position of a point with respect to the origin is represented by the position vector. For a point with coordinates (3, 4, 5):
r = 3i + 4j + 5k
The tail of the position vector lies at the origin, and its head lies at the given point.
Velocity is the rate of change of position:
v = dr/dt
For one-dimensional motion:
v = dx/dt
Acceleration is the rate of change of velocity:
a = dv/dt
The sequence is:
Position -> Velocity -> Acceleration
through differentiation. The reverse process uses integration:
Acceleration -> Velocity -> Position
Integration of velocity gives displacement or change in position.
Average velocity is based on total displacement:
Average velocity = Total displacement / Total time
Average speed is based on total distance:
Average speed = Total distance / Total time
Average acceleration is:
Average acceleration = Change in velocity / Time interval
Instantaneous velocity is the velocity at a particular instant:
v = dr/dt
Instantaneous speed is the magnitude of instantaneous velocity:
Instantaneous speed = |v|
|
Quantity |
Average Form |
Instantaneous Form |
|---|---|---|
|
Velocity |
Total displacement / Total time |
dr/dt |
|
Speed |
Total distance / Total time |
Magnitude of velocity |
|
Acceleration |
Change in velocity / Time |
dv/dt |
The slope of an x-t graph gives velocity:
Slope of x-t graph = v
The slope of a v-t graph gives acceleration:
Slope of v-t graph = a
A tangent to an x-t graph gives instantaneous velocity.
The area under a v-t graph gives displacement. Areas above and below the time axis are treated algebraically:
Displacement = Area above axis - Area below axis
For distance, all areas are added as positive magnitudes:
Distance = Area above axis + Area below axis
The area under an a-t graph gives change in velocity, not velocity directly:
Area under a-t graph = Change in velocity
The initial velocity must be added to find the final velocity.
The v-t graph is especially important because it connects slope, area, acceleration, displacement, distance, average velocity, and average speed.
If velocity and acceleration have the same sign, the particle speeds up. If they have opposite signs, the particle slows down.
The standard equations apply only when acceleration is constant:
v = u + at
s = ut + 1/2 at^2
v^2 = u^2 + 2as
For solving numerical problems:
Write the known quantities.
Identify the required quantity.
Select an equation containing the known and required values.
Substitute values with correct signs.
For variable acceleration, use differentiation or integration instead of the standard equations.
For uniformly accelerated motion, average velocity is:
Average velocity = (u + v) / 2
The same result applies to angular velocity:
Average angular velocity = (Initial angular velocity + Final angular velocity) / 2
For a body dropped from height h:
Time of fall = sqrt(2h/g)
Speed before reaching ground = sqrt(2gh)
For vertically projected particles, first choose the positive direction. Then assign signs to displacement, velocity, and acceleration.
If upward is positive, downward acceleration is negative.
If downward is positive, downward acceleration is positive.
A velocity opposite to the selected positive direction must be written as negative.
For an object released from an upward-moving balloon, the object retains the balloon’s upward velocity at the instant of release. It does not start with zero velocity.
For a projectile launched with speed u at angle theta:
Time of flight = 2u sin theta / g
Maximum height = u^2 sin^2 theta / 2g
Horizontal range = u^2 sin 2theta / g
For ground-to-ground projection, the range is maximum at an angle of 45 degrees.
Projectile motion is analysed independently along the horizontal and vertical directions:
Horizontal acceleration is zero.
Horizontal velocity remains constant.
Vertical acceleration is due to gravity.
At the highest point, vertical velocity is zero, but horizontal velocity remains present.
For a trajectory equation, such as:
y = alpha x - beta x^2
the range is found by putting y = 0:
Range = alpha/beta
The maximum height occurs at:
x = Range / 2
The general trajectory equation is:
y = x tan theta - gx^2 / (2u^2 cos^2 theta)
The velocity of A with respect to B is:
v(A/B) = v(A) - v(B)
Similarly:
a(A/B) = a(A) - a(B)
The subtraction must be vector-wise, not a subtraction of magnitudes.
Relative velocity is a vector subtraction, and relative speed is the magnitude of the resulting vector.
Two common cases are minimum-distance crossing and minimum-time crossing.
The swimmer aims upstream so that the component opposing the river current cancels it:
v sin theta = u
The swimmer reaches the point directly opposite the starting point, so downstream drift is zero.
The swimmer moves perpendicular to the river:
Time = River width / Swimmer speed
The river causes downstream drift:
Drift = River speed x Time
|
Feature |
Minimum Distance |
Minimum Time |
|---|---|---|
|
Direction |
Upstream angle |
Perpendicular to river |
|
Current cancellation |
Yes |
No |
|
Drift |
Zero |
Present |
|
Time |
Width / Effective crossing component |
Width / Swimmer speed |
For constant retardation, stopping distance is proportional to the square of initial speed:
Stopping distance = u^2 / 2a
If speed becomes one-third, stopping distance becomes one-ninth.
For a projectile, if range equals maximum height:
Range = Maximum height
then:
tan theta = 4
For minimum separation between two particles, use relative velocity. The minimum distance occurs when the separation vector becomes perpendicular to the relative velocity.
For two projectiles having the same gravitational acceleration, relative acceleration is zero. Hence, their relative path is a straight line while both remain in the air.
For a lift accelerating upward with acceleration a, effective gravity is:
Effective g = g + a
For a coin dropped inside the lift:
Time = sqrt(2h / Effective g)
Thus:
Time = sqrt(2h / (g + a))
This method is shorter than writing complete relative-motion equations.
These questions are written exactly as shown in your screenshots.
1. A ball is dropped from the top of a 100 m high tower on a planet. In the last ½ s before hitting the ground, it covers a distance of 19 m. Acceleration due to gravity (in ms⁻²) near the surface on that planet is ______. [JEE Main – 2020]
2. A Tennis ball is released from a height h and after freely falling on a wooden floor it rebounds and reaches height h/2. The velocity versus height of the ball during its motion may be represented graphically by: [JEE Main – 2020]
3. The speed verses time graph for a particle is shown in the figure. The distance travelled (in m) by the particle during the time interval t = 0 to t = 5 s will be ______.
4. A particle is projected with velocity v₀ along x-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e., ma = −αx². The distance at which the particle stops: [JEE Main – 2021]
5. An engine of a train, moving with uniform acceleration, passes the signal-post with velocity u and the last compartment with velocity v. The velocity with which middle point of the train passes the signal post is: [JEE Main – 2021]
These tips cover graphs, equations of motion, projectiles and relative motion, all taken from your document.
Speed vs velocity: Velocity uses displacement and speed uses distance. Displacement has direction, and distance does not.
Graph slopes: The slope of an x-t graph gives velocity, and the slope of a v-t graph gives acceleration.
Graph areas: The area under a v-t graph gives displacement (areas below the axis are subtracted). For distance, add all areas as positive. The area under an a-t graph gives only the change in velocity, so add the initial velocity.
Speeding up: A particle speeds up when velocity and acceleration have the same sign, and slows down when their signs are opposite.
Equations of motion: v = u + at, s = ut + ½at² and v² = u² + 2as apply only for constant acceleration. For variable acceleration, use differentiation or integration (as in Q4).
Average velocity: For uniform acceleration, it is (u + v)/2.
Sign convention. Choose the positive direction first. If upward is positive, downward acceleration is negative.
Balloon trick: An object released from an upward-moving balloon keeps the balloon’s upward velocity.
Free fall: Time of fall is √(2h/g), and speed before landing is √(2gh).
Projectile motion: Analyse x and y independently. Horizontal acceleration is zero. At the top, vertical velocity is zero, but horizontal velocity remains.
Projectile results: Range is maximum at 45° for ground-to-ground projection. If range equals maximum height, tan θ = 4. For y = αx − βx², range is α/β and maximum height occurs at x = Range/2.
Relative motion: v(A/B) = v(A) − v(B), and the subtraction is vector-wise. “A with respect to B” means observing A from B’s frame.
River crossing: For minimum time, head perpendicular to the river (drift is present). For minimum distance, use v sin θ = u (drift is zero).
Stopping distance: It equals u²/2a, so it is proportional to u². If speed becomes one-third, stopping distance becomes one-ninth.
Two projectiles: Their relative acceleration is zero, so their relative path is a straight line while both are in the air.
Lift problems: For an upward-accelerating lift, effective g = g + a, and the fall time is √(2h/(g + a)).
Kinematics rewards clear graph reading, correct sign conventions and vector thinking. Master slopes and areas, equations of motion, projectile components, relative velocity and river crossing, then practise the PYQs above until each pattern feels automatic. Revise these tips, and Kinematics becomes one of the safest scoring areas in NEET 2027.
