Motion in a Straight Line

Motion in a Straight LineNEET Physics · Class 11 · NCERT Chapter 2

Overview Notes NEET Questions Interactive Learning

8 interactive concept widgets for Motion in a Straight Line. Drag any slider, change any number, and watch the formula and the answer update live. Built so you understand how each NEET problem actually works, not just the final number.

On this page

Kinematic equation solver

Distance in the n-th second

Free fall simulator (drop or throw up)

Stopping distance calculator

V-T graph: area = displacement

P-T graph reader (slope = velocity)

Relative velocity (1D) and crossing time

Average speed vs average velocity

Kinematic equations

The three equations of motion are the engine of every NEET problem in this chapter. Pick an equation, fill in the values you know, and watch the missing variable fall out.

Kinematic equations

Kinematic equation solver

Pick one of the three equations of motion, fill in the values you know, and the solver finds the missing variable with the full step shown.

v = u + a t

Initial velocity u

m/s

Final velocity v

m/s

Acceleration a

m/s²

Time t

Answer

v = 10.000 m/s

v = u + a t = 0 + (2) (5) = 10.00 m/s

Try this

Leave exactly one field blank — the solver fills it in.
Switch equations to see why some problems need v² = u² + 2as (no time given).
Try negative acceleration to model braking (deceleration).

Kinematic equations

Distance in the n-th second

A common NEET trap: the displacement during a single second is not the same as the cumulative displacement at that time. Move the sliders to see both.

Initial velocity u: 0 m/s

Acceleration a: 4 m/s²

Which second n: 5

Formula

s_{n} = u + \frac{a}{2} (2 n - 1) = 0 + \frac{4}{2} (2 \times 5 - 1) = 18.00 m

Total dist at t=4s

32.00 m

Total dist at t=5s

50.00 m

Distance in nth s

18.00 m

Try this

Set u = 0, a = 10 m/s², watch the n-th second distances form an arithmetic series: 5, 15, 25, 35... (Galileo's law of odd numbers).
For a body in free fall (a = g = 10), the distance in the 1st, 2nd, 3rd second is in ratio 1:3:5.
Increase u: the n-th second distance starts higher but the gap between consecutive seconds stays the same (it equals a).

Free fall and stopping distance

Free fall is the most-tested special case of constant acceleration. Stopping distance shows the famous u² scaling.

Free fall under gravity

Free fall simulator

Toggle between dropping a stone and throwing one upward. Adjust the slider to see how time, speed and max height scale.

Drop height h: 80 m

Time to ground

4.00 s

Impact speed

40.00 m/s

Working

t = \frac{2 h}{g} = \frac{2 \times 80}{10} = 4.00 s

v = g t = 10 \times 4.00 = 40.00 m/s

Using g = 10 m/s², ignoring air resistance.

Try this

Drop from 80 m → 4 s to reach the ground (the classic NEET answer).
Throw up at 20 m/s → max height 20 m, total flight 4 s. The way up and way down each take 2 s.
Halve the throw speed and the max height drops to a quarter — h ∝ u², so this is the same scaling as stopping distance.

Free fall under gravity

Stopping distance calculator

A vehicle moving with speed u brakes to rest with deceleration a. Move the sliders to see how stopping distance scales with the square of the speed.

Initial speed u: 20 m/s (72 km/h)

Deceleration a: 5 m/s²

Stopping distance

s = 40.00 m

s = \frac{u ^{2}}{2 a} = \frac{2 0 ^{2}}{2 \times 5} = 40.00 m

t = \frac{u}{a} = \frac{20}{5} = 4.00 s

Doubling the speed quadruples the distance. At 40 m/s on the same road, you would need 160.00 m to stop — exactly 4× the current value, because s ∝ u².

Try this

Set u = 20 m/s, a = 5 m/s² → 40 m. Now set u = 30 m/s → 90 m. Same road, 50% more speed, 125% more stopping distance.
NEET tip: examiners almost always test the u² scaling. If the speed in the question is doubled or tripled, square it.
Wet roads have lower friction (lower a). Halving a doubles the stopping distance.

Position-time and velocity-time graphs

Read graphs the way NEET tests them. Slopes are velocities and accelerations; areas are displacements.

Graphs

V-T graph: area = displacement

Move the sliders to change u, v and t. The trapezium fills the area under the velocity-time graph — that area equals the displacement.

Initial velocity u: 10 m/s

Final velocity v: 30 m/s

Duration t: 8 s

Displacement = area under v-t graph

s = 160.00 m

s = \frac{1}{2} (u + v) t = \frac{1}{2} (10 + 30) (8) = 160.00 m

Slope of the line = acceleration = 2.50 m/s²

Try this

Set u = 10, v = 30, t = 8 → trapezium with two parallel sides (10 and 30), height 8 → area 160 m.
Set u = v → rectangle (uniform velocity). Slope is zero, no acceleration.
Set v < u → object decelerating. Trapezium tilts down; area still gives the displacement.
Set u = 0, v > 0 → triangle. Area = ½ × base × height = ½ × t × v.

Graphs

Position-time graph reader

Pick a motion type and slide the cursor to read the position and the instantaneous velocity (the slope of the tangent) at any instant.

Cursor at t = 4.0 s

Position x(t)

12.00 m

Velocity = slope

6.00 m/s

v = \frac{d x}{d t}_{t = 4.0} = 6.00 m/s

The orange dashed line is the tangent at the cursor. Its slope equals the instantaneous velocity.

Try this

On the "At rest" graph, the slope is zero everywhere — velocity is zero at every t.
On the "Uniform v" graph, the slope is constant — velocity does not change with time.
On the "Accelerating" parabola, slide the cursor and watch the slope grow linearly with t.
On the "Decelerating" curve, the slope decreases until it crosses zero (the moment the body stops).

Relative velocity and averages

One-dimensional relative velocity sets the time to overtake or cross. The round-trip example shows why average speed and average velocity are not the same.

Relative velocity

Relative velocity (one dimension)

Two objects on a straight line. Pick directions and speeds to see v_AB and the time to overtake or cross.

Velocity of A (v_A): 20 m/s

Speed of B (|v_B|): 10 m/s

Relative velocity of A w.r.t. B

v_AB = 10 m/s

v_{A B} = v_{A} - v_{B} = 20 - (10) = 10 m/s

In the same direction, you subtract magnitudes. Negative means B is faster.

Crossing/overtaking time (treating A and B as trains)

Length of A: 100 m

Length of B: 200 m

Time to overtake = 30.00 s (300 m / 10 m/s)

Try this

Two trains moving the same direction at 20 and 10 m/s: relative velocity 10 m/s. Lengths 100 + 200 m → 30 s to overtake.
Same trains moving opposite: relative velocity 30 m/s, only 10 s to cross.
If both have the same speed in the same direction, v_AB = 0 — A never overtakes B.

Velocity

Average speed vs average velocity

Average speed and average velocity are different. The classic round trip example shows why — and why the harmonic mean (not arithmetic) is the correct answer.

A car drives d km away at v₁, then drives back the same distance at v₂. Watch the difference between average speed and average velocity.

One-way distance d: 60 km

Outbound speed v₁: 60 km/h

Return speed v₂: 40 km/h

Average speed

48.00 km/h

Average velocity

0 km/h

net displacement = 0

Working

\overset{v}{ˉ}_{speed} = \frac{2 d}{d / v _{1} + d / v _{2}} = \frac{2 v _{1} v _{2}}{v _{1} + v _{2}} = \frac{2 \times 60 \times 40}{60 + 40} = 48.00 km/h

Note this is the harmonic mean (48.00), not the arithmetic mean (50.00). NEET tests this distinction every other year.

Try this

Set v₁ = 60, v₂ = 40 → average speed = 48 km/h, not 50. The harmonic mean is always smaller than the arithmetic mean.
Set v₁ = v₂ → both means equal that speed (the only case where average = both speeds).
No matter what v₁, v₂ you pick, the average velocity for the round trip is always 0.

Next chapter interactive widgets

Drag, slide and recompute on the next chapter's widgets.

Previous chapter

Class 11 · Ch 1

Units and Measurements

Open interactive widgets

Next chapter

Class 11 · Ch 3

Motion in a Plane

Open interactive widgets

Track Your NEET Score Across All 90 Chapters

Free 14-day trial. AI tutor, full mock tests and chapter analytics — built for NEET 2027.

Free 14-day trial · No credit card required