Motion in a Straight Line

1. Path Length and Displacement

Motion in 1D involves change in position along a straight line.

Distance (Path Length): Total path covered. Scalar. Always positive.
Displacement: Change in position. Vector. Can be positive, negative, or zero.
Average Speed: Total Distance / Total Time.
Average Velocity: Total Displacement / Total Time.

2. Kinematic Equations

Valid only for Uniform Acceleration (constant a).

1. v = u + at
2. s = ut + ½at²
3. v² = u² + 2as
4. s_n = u + (a/2)(2n - 1) (Displacement in nth second)

(Asked in NEET 2016, 2019, 2021)

3. Motion Under Gravity

Free fall under gravity (a = -g).

Time to Reach Max Height: t = u/g.
Maximum Height: H = u²/2g.
Time of Flight: T = 2u/g.
Galileo's Ratio: Distances derived in equal time intervals are in ratio 1:3:5:7...

4. Relative Velocity

v_AB = v_A - v_B.
Same Direction: |v_A - v_B|.
Opposite Direction: |v_A + v_B|.

(Asked in NEET 2017, 2020, 2022)

Numericals: Motion in 1D

Q1. A car covers first half of distance at 40 km/h and second half at 60 km/h. Find average speed.

Solution:
For equal distances, v_avg = 2v₁v₂ / (v₁ + v₂).
v_avg = (2 × 40 × 60) / (40 + 60)
v_avg = 4800 / 100 = 48 km/h.

Q2. A car moving at 20 m/s stops in 40 m on applying brakes. If it moves at 40 m/s, what is stopping distance (assuming same retardation)?

Solution:
d = u² / 2a. So d ∝ u².
d₂ / d₁ = (u₂ / u₁)²
d₂ / 40 = (40 / 20)² = 2² = 4.
d₂ = 40 × 4 = 160 m.

Q3. A body starts from rest with a = 2 m/s². Find distance traveled in 4th second.

Solution:
s_n = u + (a/2)(2n - 1).
u = 0, a = 2, n = 4.
s₄ = 0 + (2/2)(2×4 - 1) = 1 × 7 = 7 m.

Q4. A ball dropped from a tower reaches ground in 4 s. Find height of tower. (g=10)

Solution:
h = ut + ½gt². (u=0)
h = 0 + 0.5 × 10 × (4)²
h = 5 × 16 = 80 m.

Q5. Two trains 100 m and 200 m long move at 60 km/h and 90 km/h in opposite directions. Time to cross?

Solution:
Total Distance D = 100 + 200 = 300 m = 0.3 km.
Relative Speed V = 60 + 90 = 150 km/h (Opposite).
Time t = D/V = 0.3 / 150 hr.
t = (0.3/150) × 3600 sec = 0.3 × 24 = 7.2 s.

Q6. Driver sees hazard, reaction time 0.2s. Brakes cause retardation 5 m/s². Speed 10 m/s. Stopping dist?

Solution:
Dist during reaction = u × t_r = 10 × 0.2 = 2 m.
Braking distance = u² / 2a = 100 / (2×5) = 10 m.
Total = 2 + 10 = 12 m.

Q7. Ball thrown up returns in 6 s. Find initial velocity. (g=10)

Solution:
Total time T = 2u/g.
6 = 2u / 10.
2u = 60 → u = 30 m/s.

Q8. Displacement x = 3t² + 2t. Find velocity at t=2s.

Solution:
v = dx/dt.
v = 6t + 2.
At t=2: v = 6(2) + 2 = 12 + 2 = 14 m/s.

Q9. Velocity v = √(180 - 16x). Find acceleration.

Solution:
Square both sides: v² = 180 - 16x.
Compare with v² = u² + 2as.
2a = -16.
a = -8 m/s². (Retardation).

Q10. A juggler throws balls every 2 s. How high must they go so 2 balls are in air?

Solution:
For more than 1 ball in air, time of flight > interval.
Let ball 1 be landing when ball 3 is thrown. Time of flight = 2 * interval = 4 s.
T = 4 = 2u/g → u = 20 m/s.
H = u²/2g = 400 / 20 = 20 m.

Important Formulae

1. Equations of Motion

Uniform Acceleration:

v = u + at

s = ut + ½at²

v² - u² = 2as

Distance in nth second: s_n = u + (a/2)(2n - 1)

2. Free Fall

Dropped from height h:

t = √(2h/g) | v = √(2gh)

Thrown up with u:

H_max = u²/2g | T_flight = 2u/g

3. Special Cases

Stopping Distance:

d_s = u² / 2a

Reaction Time:

Total Dist = (v × t_r) + (v²/2a)

4. Relative Velocity

v_AB = v_A - v_B

Overtaking Time: Distance / Relative Speed
Crossing Time: (L₁+L₂) / Relative Speed

20 NEET Golden Facts

1. Distance vs Displacement: Distance ≥ |Displacement|. Equal only for motion in straight line without turning back.
2. Average Velocity: Depends only on start and end points. Can be zero for round trip.
3. Instantaneous Speed: Magnitude of instantaneous velocity vector. Always positive.
4. Slope of x-t Graph: Gives velocity. Slope of v-t graph gives acceleration.
5. Area under v-t Graph: Gives displacement (with sign) or distance (if sum of absolute areas).
6. Area under a-t Graph: Gives Change in velocity (Δv).
7. Free Fall: Near earth surface, acceleration is constant (g). Equations of motion apply.
8. Galileo's Ratio: Distances in equal intervals t starting from rest: 1 : 3 : 5 : 7.
9. Max Height time: Time to go up = Time to come down (Air resistance neglected).
10. Air Resistance: If considered, Time of descent > Time of ascent. Acceleration down (g-a), Deceleration up (g+a).
11. Relative Velocity: Object A appears to move with v_A - v_B for observer B.
12. Constant Acceleration: Velocity changes linearly with time. Position changes quadratically.
13. Crossing Trains: Total distance is sum of lengths. Relative speed is sum (opp direction) or diff (same direction).
14. Zero Velocity: A body can have zero velocity but non-zero acceleration (e.g., at top of flight).
15. Mid-Point Speed: If speed at A is u and B is v, speed at mid-point is √[(u²+v²)/2].
16. Retardation: Negative acceleration (Speed decreases).
17. Variable Accel: Need calculus. v = ∫a dt, x = ∫v dt.
18. Speed-Time Graph: Cannot be negative.
19. Distance in last second: Drop from height H, distance in last second is same as distance in 1st second of upward throw with impact velocity.
20. Average Speed (Half Distance): v_avg = 2v₁v₂ / (v₁+v₂). (Harmonic Mean).

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