Motion in a Straight Line

Correct Answer: B

Kinematics describes motion — position, velocity, acceleration, displacement — without inquiring into what causes the motion. Dynamics (Newton's Laws, Chapter 4) deals with causes.

Correct Answer: C — Slope of the tangent at that point

v = dx/dt = limit of Δx/Δt as Δt → 0. Graphically, this limit is the slope of the tangent to the x-t curve at that instant. Average velocity is the slope of a chord.

Correct Answer: C — Displacement

The area under v-t graph = ∫v dt = displacement. For uniform velocity u over time T: area = u × T = uT = displacement. Note: distance = area only when velocity doesn't change sign.

Correct Answer: C — A parabola

x = v₀t + ½at² is a quadratic in t → parabola. For uniform motion (a=0), x = v₀t is linear → straight line inclined to time axis. The v-t graph for uniform acceleration is a straight line.

Correct Answer: C — 9.8 m s⁻² downward

Zero velocity does NOT imply zero acceleration. Acceleration due to gravity (g = 9.8 m s⁻²) acts downward throughout the motion — at the highest point, on the way up, and on the way down.

Correct Answer: C — Instantaneous acceleration

a = dv/dt. On a v-t graph, this is the slope of the tangent at that point. Average acceleration is the slope of the chord (secant) between two points.

Correct Answer: B — v₀²

Using v² = v₀² + 2ax with v = 0: 0 = v₀² − 2a·dₛ → dₛ = v₀²/(2a). Thus, stopping distance ∝ v₀². Doubling speed increases stopping distance by 4 times!

Correct Answer: B — v̄ = (v₀ + v) / 2

For constant acceleration only: v̄ = (v₀ + v)/2. This is the arithmetic mean of initial and final velocities. This formula is NOT valid for non-uniform acceleration.

Correct Answer: C — 10.0 m s⁻¹

x = a + bt² → v = dx/dt = 2bt
At t = 2s: v = 2 × 2.5 × 2 = 10 m s⁻¹

Correct Answer: C — 1 : 3 : 5 : 7 ...

Using y = ½gt², positions after τ, 2τ, 3τ... are y₀, 4y₀, 9y₀... The distances traversed in successive intervals are y₀, 3y₀, 5y₀, 7y₀... → ratio 1:3:5:7 (odd numbers). Galileo established this experimentally.

Correct Answer: B

At any instant: speed = |velocity|. Average speed ≥ |average velocity| (equality only for straight-line motion in one direction). But instantaneous speed = |instantaneous velocity| always.

Correct Answer: B — Speed is decreasing

If velocity is positive (+ve direction) and acceleration is negative (opposite to velocity), the particle is decelerating — its speed is decreasing. It still moves in the positive direction (until v = 0).

Correct Answer: C — y = −½gt²

Taking upward as positive: a = −g, v₀ = 0. x = v₀t + ½at² = 0 + ½(−g)t² = −½gt². The negative sign confirms the object falls downward (negative direction).

Correct Answer: B — v_A − v_B

Relative velocity of A with respect to B: v_AB = v_A − v_B. For two objects moving in the same direction, relative speed = |v_A − v_B|. For opposite directions, relative speed = v_A + v_B.

Correct Answer: A — 105 m s⁻¹

Van speed = 30 km/h = 30/3.6 = 8.33 m/s
Bullet speed (ground) = 150 + 8.33 = 158.33 m/s
Thief's car = 192 km/h = 192/3.6 = 53.33 m/s
Relative speed of bullet w.r.t. car = 158.33 − 53.33 = 105 m s⁻¹

Correct Answer: B

Instantaneous speed = |instantaneous velocity| at every instant. They are always equal in magnitude. In contrast, average speed ≥ |average velocity|, and displacement ≤ distance.

Correct Answer: C — The particle is changing direction

Velocity is a vector (magnitude + direction). Constant speed means |v| = constant, but if direction changes, velocity changes. This is the case in uniform circular motion, for example.

Correct Answer: B — 35 m s⁻¹

Let A move in +ve direction (v_A = +20), B in −ve direction (v_B = −15).
Relative velocity of A w.r.t. B = v_A − v_B = 20 − (−15) = 35 m s⁻¹

Correct Answer: B — Only for constant acceleration

These kinematic equations are derived assuming acceleration is constant throughout. For variable acceleration, calculus must be used: v = ∫a dt and x = ∫v dt.

Correct Answer: B — 0.2 s

d = ½gt² → t = √(2d/g) = √(2×0.21/9.8) = √(0.0429) ≈ 0.207 s ≈ 0.2 s

Correct Answer: B — 1 : 3 : 5

Using Galileo's law of odd numbers (Sₙ = ½a(2n−1)):
S₁ = ½a(1) = a/2
S₂ = ½a(3) = 3a/2
S₃ = ½a(5) = 5a/2
Ratio = 1 : 3 : 5 ✓

Correct Answer: C — 29.4 m s⁻¹

v = v₀ + gt = 0 + 9.8 × 3 = 29.4 m s⁻¹

Correct Answer: C — Positive acceleration

An x-t graph curving upward (concave up) means slope (velocity) is increasing with time → positive acceleration. Curving downward (concave down) means negative acceleration. A straight line means zero acceleration.

Correct Answer: C — Greater than or equal to

Average speed = total path length / time. |Average velocity| = |displacement| / time. Since path length ≥ |displacement|, average speed ≥ |average velocity|. Equality holds only when motion is in one direction without reversal.

Correct Answer: B — 20 m

At max height v = 0. Using v² = v₀² − 2gh:
0 = (20)² − 2(10)(h) → h = 400/20 = 20 m

Instantaneous velocity: v = dx/dt = d(8.5 + 2.5t²)/dt = 5.0t m s⁻¹ (a) At t = 0: v = 5.0 × 0 = 0 m s⁻¹

(b) At t = 2 s: v = 5.0 × 2 = 10 m s⁻¹

(c) Average velocity (t = 2 to 4 s): x(4) = 8.5 + 2.5(16) = 8.5 + 40 = 48.5 m x(2) = 8.5 + 2.5(4) = 8.5 + 10 = 18.5 m v̄ = [x(4) − x(2)] / (4−2) = 30/2 = 15 m s⁻¹ Average velocity = 15 m s⁻¹

Take upward as +ve. Launch point y₀ = 25 m, v₀ = +20 m/s, a = −10 m/s²

(a) Max height above launch point: v² = v₀² + 2a(y − y₀) 0 = 400 + 2(−10)(y − 25) y − 25 = 20 m → y = 45 m above ground Max height above ground = 45 m

(b) Total time (using quadratic — Second Method): 0 = 25 + 20t + ½(−10)t² 5t² − 20t − 25 = 0 t² − 4t − 5 = 0 → (t−5)(t+1) = 0 t = 5 s (taking positive root)
Total time = 5 s

v₀ = 126 km/h = 126/3.6 = 35 m/s, v = 0, x = 200 m

(a) Retardation: v² = v₀² + 2ax 0 = (35)² + 2a(200) a = −1225/400 = −3.0625 m s⁻² Retardation = 3.06 m s⁻² ≈ 3.1 m s⁻²

(b) Time to stop: v = v₀ + at → 0 = 35 + (−3.0625)t t = 35/3.0625 ≈ 11.4 s Time = 11.4 s

v₀ = 29.4 m/s, v = 0 at top, a = −9.8 m/s²

(a) Maximum height: v² = v₀² − 2gh 0 = (29.4)² − 2(9.8)h h = 864.36/19.6 = 44.1 m Maximum height = 44.1 m

(b) Time to return:
Time to reach top: t₁ = v₀/g = 29.4/9.8 = 3 s
By symmetry, time to fall back = 3 s Total time = t₁ + t₂ = 3 + 3 = 6 s Total time = 6 s

Time going = 2.5/5 = 0.5 h = 30 min
Time returning = 2.5/7.5 = 1/3 h = 20 min
Total time = 50 min = 5/6 h

(a) Average velocity (complete journey): Displacement = 0 (returns to start) |Average velocity| = 0/t = 0 m s⁻¹ |Average velocity| = 0 km h⁻¹

(b) Average speed: Total path = 2.5 + 2.5 = 5 km Average speed = 5 / (5/6) = 6 km h⁻¹ Average speed = 6 km h⁻¹
(This illustrates why average speed ≠ |average velocity|!)

Method 1 — Calculus: v = dx/dt = d(0.08t³)/dt = 0.24t² At t = 4: v = 0.24 × 16 = 3.84 m s⁻¹ ✓ Method 2 — Limiting process (Δt → 0): Δx/Δt for small Δt centred at t = 4: Δt = 0.5: x(4.25)−x(3.75) / 0.5 = 0.08(4.25³−3.75³) / 0.5 ≈ 3.845 Δt = 0.1: ≈ 3.8402 Δt→0: approaches 3.84 m s⁻¹ Instantaneous velocity at t = 4 s = 3.84 m s⁻¹

Speed just before 1st bounce: v² = 2gh = 2 × 10 × 90 = 1800 v = √1800 = 30√2 ≈ 42.43 m s⁻¹ Speed just after 1st bounce: v' = v − (1/10)v = (9/10)v = 0.9 × 42.43 ≈ 38.18 m s⁻¹ Height after 1st bounce: h' = v'² / (2g) = (38.18)² / 20 = 1458/20 = 72.9 m Height after 1st bounce = 72.9 m ≈ (0.81 × 90 m)

Each cycle (8 steps, 8 s): net displacement = 5 − 3 = +2 m

After 1 cycle (8s): position = 2 m After 2 cycles (16s): position = 4 m After 3 cycles (24s): position = 6 m After 4 cycles (32s): position = 8 m After 5 cycles (40s): position = 10 m After 5 complete cycles: 10 m. Now 5 more steps forward:
Step 1: 11 m, Step 2: 12 m, Step 3: 13 m ← Falls in pit! Total time = 40 s + 3 s = 43 s Time to reach pit = 43 s

Moving in opposite directions, relative speed = v_A + v_B v_rel = 60 + 80 = 140 km h⁻¹ Time = Distance / v_rel = 350 / 140 = 2.5 h Time = 2.5 hours = 2 h 30 min

v₀ = 0, a = g = 9.8 m/s², h = 78.4 m

(a) Time: h = ½gt² → 78.4 = ½ × 9.8 × t² t² = 156.8/9.8 = 16 → t = 4 s Time = 4 s

(b) Final velocity: v = gt = 9.8 × 4 = 39.2 m s⁻¹ Velocity = 39.2 m s⁻¹ (downward)

v = 72 km/h = 20 m/s

Phase 1 (acceleration, 0 to 20 m/s in 10 s): d₁ = ½(v₀+v)t = ½(0+20)(10) = 100 m Phase 2 (constant speed, 5 min = 300 s): d₂ = v × t = 20 × 300 = 6000 m Phase 3 (deceleration, 20 to 0 in 20 s): d₃ = ½(20+0)(20) = 200 m Total distance: d = 100 + 6000 + 200 = 6300 m Total distance = 6300 m = 6.3 km

(a) Time to reach office: t = d/v = 2.5/5 = 0.5 h = 30 min Reaches at 9:30 AM (b) Return journey: t = 2.5/25 = 0.1 h = 6 min Reaches home at 5:06 PM (c) Average speed for whole day (travel only): Total path = 2.5 + 2.5 = 5 km Travel time = 0.5 + 0.1 = 0.6 h Avg speed = 5/0.6 = 8.33 km h⁻¹ Average travel speed = 8.33 km h⁻¹
Note: If total time including office hours (8.5h) is used, avg speed = 5/9.1 ≈ 0.55 km/h

Derivation:
Final velocity v = 0, initial velocity = v₀, retardation = a
Using v² = v₀² − 2a·dₛ: 0 = v₀² − 2a·dₛ dₛ = v₀² / (2a) Stopping distance dₛ = v₀² / (2a)

Effect of doubling speed: dₛ' = (2v₀)² / (2a) = 4v₀² / (2a) = 4dₛ Stopping distance increases by a factor of 4
This is critical for road safety — doubling speed is far more dangerous than it seems!

v₀ = 0, a = 4 m s⁻²

(a) Velocity at t = 5 s: v = v₀ + at = 0 + 4(5) = 20 m s⁻¹ v = 20 m s⁻¹

(b) Distance in 5 s: x = v₀t + ½at² = 0 + ½(4)(25) = 50 m x = 50 m

(c) Distance in 5th second (Sₙ = v₀ + a(n − ½)): S₅ = v₀ + a(5 − ½) = 0 + 4 × 4.5 = 18 m Distance in 5th second = 18 m
Verify: S₄ = ½(4)(16) = 32 m; S₅ − S₄ = 50 − 32 = 18 m ✓

v_B = 54 km/h = 15 m/s (constant), v_A₀ = 0, a_A = 2 m/s²

At time t, positions (A starts from 0, B starts from 200 m): x_A = ½ × 2 × t² = t² x_B = 200 + 15t A catches B when x_A = x_B: t² = 200 + 15t t² − 15t − 200 = 0 t = [15 + √(225 + 800)] / 2 = [15 + √1025] / 2 = [15 + 32.02] / 2 ≈ 23.5 s Distance A travels: x_A = t² = (23.5)² ≈ 552 m A catches B after t ≈ 23.5 s, having travelled ≈ 552 m