Motion in a Straight Line

10 s

Step 1: Let t be the time from start of A when B overtakes. B starts at t = 5 s, so B travels for time (t – 5) s. Step 2: Position of A: x_A = ½ × 2 × t² = t². Step 3: Position of B: x_B = 20(t – 5) [B starts at t = 5 s with constant velocity 20 m/s]. Step 4: At overtaking, x_A = x_B: t² = 20(t – 5) = 20t – 100. Step 5: t² – 20t + 100 = 0. Using quadratic formula: t = [20 ± √(400 – 400)]/2 = [20 ± √0]/2. Discriminant = 400 – 400 = 0, so t = 10 s. Wait — this gives t = 10 s as the only solution (tangent point, not overtaking). For overtaking to happen, we need x_B > x_A at some point. Check: at

Twice – once while going up and once while coming down

Step 1: Maximum height reached: H = u²/2g = (19.6)²/(2×9.8) = 384.16/19.6 = 19.6 m. Step 2: Since 14.7 m < 19.6 m (maximum height), the ball does cross this height. Step 3: Using h = ut – ½gt²: 14.7 = 19.6t – ½(9.8)t² → 14.7 = 19.6t – 4.9t². Step 4: 4.9t² – 19.6t + 14.7 = 0 → t² – 4t + 3 = 0 → (t–1)(t–3) = 0 → t = 1 s and t = 3 s. Step 5: Two different positive values of t mean the ball is at 14.7 m twice: at t = 1 s (going up) and t = 3 s (coming down). The ball reaches maximum height at t = u/g = 19.6/9.8 = 2 s, confirming t=1s is upward journey and t=3s is downward journey.

Positive velocity and negative acceleration (retardation)

Step 1: The slope of an x-t (displacement-time) graph gives the instantaneous velocity. Step 2: Since displacement is positive and increasing, the slope (velocity) is positive. Step 3: Since the graph is concave down (bends downward), the slope is decreasing with time. Step 4: A decreasing slope means velocity is decreasing — this implies negative acceleration (retardation). Step 5: Summary: Positive displacement → particle is ahead of reference point. Positive but decreasing slope → positive velocity that is reducing, i.e., the particle is slowing down while still moving in the positive direc

v_max = a₁a₂T/(a₁ + a₂)

Step 1: Let t₁ be the time to accelerate and t₂ be the time to decelerate. Then t₁ + t₂ = T. Step 2: During acceleration phase: v_max = a₁t₁ → t₁ = v_max/a₁. Step 3: During deceleration phase: 0 = v_max – a₂t₂ → t₂ = v_max/a₂. Step 4: Adding: t₁ + t₂ = v_max/a₁ + v_max/a₂ = v_max(a₁ + a₂)/(a₁a₂) = T. Step 5: Solving: v_max = a₁a₂T/(a₁ + a₂). This is the harmonic mean relationship. Common error: Students often use arithmetic average (a₁+a₂)T/2, which is dimensionally correct but physically wrong.