Beyond the Starting Line: Mastering 1D Kinematics

Motion is the heartbeat of physics. But here’s the kicker: nothing in the universe is actually “at rest.” You’re currently sitting on a planet spinning at 1,600 km/h, orbiting a sun at 107,000 km/h.

In Motion in a Straight Line, we strip away the complexity of the cosmos and focus on one thing: a point mass moving along a single axis. It sounds simple, but once you mix in variable acceleration and relative frames, the math gets spicy.

The Core Pillars of Motion

1. The Reference Frame Trap

Motion is relative. If you’re on a train moving at 60 km/h and throw a ball forward at 10 km/h, to someone on the platform, that ball is a 70 km/h projectile. To you, it’s just a slow toss. Always define your Origin and Direction (+ve or -ve) before you touch a calculator.

2. The Calculus Connection

In the real world, acceleration isn’t always constant.

• Velocity (v) is the rate of change of displacement: v = dx/dt.
• Acceleration (a) is the rate of change of velocity: a = dv/dt = d²x/dt².If you see an equation with t³ or t⁴, your standard kinematic equations won’t save you—you’ll need to differentiate or integrate.

3. Graphing the Truth

Graphs are the cheat codes of kinematics:

• Position-Time (x-t): Slope = Velocity.
• Velocity-Time (v-t): Slope = Acceleration; Area Under Curve = Displacement.
• Acceleration-Time (a-t): Area Under Curve = Change in Velocity.

The Gauntlet: 10 Challenging Aptitude Questions

Test your grit with these application-based problems. No “plug-and-play” formulas here—you’ll need to think.

Question 1: The Non-Constant Accelerator

A particle moves along the x-axis such that its velocity varies with time as v = 3t² – 6t (where v is in m/s and t is in seconds). Find the distance traveled by the particle between t = 0 and t = 3 seconds. (Careful: Displacement and Distance are not the same here!)

Question 2: The Reaction Time Disaster

A driver is traveling at 108 km/h. Suddenly, they see a roadblock 60m away. If their reaction time is 0.5 seconds and the maximum deceleration of the car is 10 m/s², will the car hit the roadblock? If so, at what speed?

Question 3: The Meeting of the Stones

Stone A is dropped from a tower of height H. At the same instant, Stone B is thrown upward from the ground with velocity u. At what time t will they pass each other, and what is the minimum value of u required for them to meet?

Question 4: The Ratio of Seconds

A body starts from rest and moves with uniform acceleration. What is the ratio of the distance covered in the 5th second of its motion to the total distance covered in 5 seconds?

Question 5: The Chasing Police

A thief’s car passes a police station at a constant speed of 20 m/s. Exactly 5 seconds later, a police jeep starts from rest and accelerates at 4 m/s². How far from the station will the police jeep overtake the thief?

Question 6: The Variable Gravity Simulation

In a sci-fi experiment, a ball is thrown upward where the “effective acceleration” is not constant but varies as a = -g – kv, where v is velocity and k is a constant. Describe qualitatively: will the time taken to reach the peak be more or less than the time taken to fall back to the start?

Question 7: The Overtaking Limit

Car A is moving at 10 m/s and Car B is 100m behind it, moving at 20 m/s. If Car A suddenly starts accelerating at 2 m/s², will Car B ever catch up? If so, when?

Question 8: The Elevator Throw

An elevator is moving upward with a constant acceleration of 2 m/s². A person inside the elevator drops a coin from a height of 2m relative to the elevator floor. How long does it take for the coin to hit the floor? (Take g = 10 m/s²)

Question 9: The Average Speed Trap

A particle covers half of its total distance with speed v₁. The rest of the distance is covered in two equal time intervals with speeds v₂ and v₃. Find the average speed for the entire journey.

Question 10: The Galilean Oddity

Show that for an object falling from rest, the distances traveled in successive equal intervals of time follow the ratio 1 : 3 : 5 : 7… (This is known as Galileo’s Law of Odd Numbers).

Detailed Explanations & Solutions

1. The Distance vs. Displacement Trap

Velocity v = 3t(t – 2). It is negative from t=0 to t=2 (moving backward) and positive from t=2 to t=3.

• Integrate |v| dt.
• ∫(0 to 2) |3t² – 6t| = 4m.
• ∫(2 to 3) (3t² – 6t) = 4m.Result: Total distance = 8m. (Displacement would be 0m).

2. Reaction Time

Initial speed u = 108 km/h = 30 m/s.

During reaction time (0.5s), distance = 30 × 0.5 = 15m.

Remaining distance = 60 – 15 = 45m.

Stopping distance required: v² = u² + 2as → 0 = 30² + 2(-10)s → s = 45m.

Result: The car stops exactly at the roadblock. Speed = 0 m/s.

3. The Meeting Stones

Let them meet at time t.

Stone A: y₁ = H – (1/2)gt².

Stone B: y₂ = ut – (1/2)gt².

Set y₁ = y₂ → H = ut → t = H/u.

For them to meet, t must be less than or equal to the time Stone B takes to reach its peak (u/g) or return. Practically, u ≥ √(gH/2).

4. The Ratio of Seconds

Distance in nth second: Sn = u + a/2(2n – 1). Since u=0, S₅ = (a/2)(9).

Total distance in 5s: Sₜ = (1/2)a(5)² = (a/2)(25).

Result: Ratio = 9 : 25.

5. Police Chase

Thief distance: x = 20(t + 5).

Police distance: x = (1/2)(4)t² = 2t².

Set 2t² = 20t + 100 → t² – 10t – 50 = 0. Solve for t ≈ 13.66s.

Distance = 2 × (13.66)² ≈ 373m.

6. Variable Gravity

When going up, both gravity and air-like resistance (kv) act downward (acceleration = -g – kv). When coming down, gravity is down but resistance is up (acceleration = -g + kv).

Result: The “upward” deceleration is stronger than the “downward” acceleration. Thus, Time of Ascent < Time of Descent.

7. Overtaking Limit

Relative velocity u_rel = 10 m/s. Relative acceleration a_rel = -2 m/s².

Max distance Car B can “gain” is when relative velocity becomes zero: v² = u² + 2as → 0 = 10² + 2(-2)s → s = 25m.

Since they were 100m apart and B only gains 25m, Car B will never catch up.

8. The Elevator Throw

Use relative acceleration.

a_coin_rel_elevator = a_coin – a_elevator = (-10) – (+2) = -12 m/s².

Distance s = -2m.

-2 = 0 + (1/2)(-12)t² → t² = 1/3.

Result: t = 1/√3 seconds ≈ 0.577s.

9. Average Speed

Let total distance be 2x. First half time t₁ = x/v₁.

Second half: distance x = v₂(t/2) + v₃(t/2) → t₂ = 2x / (v₂ + v₃).

Avg Speed = Total Dist / Total Time = 2x / [x/v₁ + 2x/(v₂ + v₃)].

Result: 2v₁(v₂ + v₃) / (v₂ + v₃ + 2v₁).

10. Galilean Odd Numbers

Distance in t intervals: (1/2)at², (1/2)a(2t)², (1/2)a(3t)²… which are 1, 4, 9, 16… Distances in successive intervals are (4-1), (9-4), (16-9)…

Result: 1 : 3 : 5 : 7…