2017 Date Name: Suresh Rane Roll No: 25 Subject: Physics Chapter: Kinematics Q1. A car accelerates from rest at a constant rate of 2 m/s² for 10 seconds. Calculate the final velocity and distance covered. Given: Initial velocity (u) = 0 m/s Acceleration (a) = 2 m/s² Time (t) = 10 s To find: Final velocity (v) = ? Distance covered (s) = ? Solution: Using the first equation of motion: v = u + at v = 0 + (2 × 10) v = 20 m/s Using the second equation of motion: s = ut + rac{1}{2}at² s = (0 × 10) + rac{1}{2}(2)(10)² s = 0 + rac{1}{2}(2)(100) s = 100 m Answer: Final velocity = 20 m/s Distance covered = 100 m --- Q2. A ball is thrown vertically upwards with an initial velocity of 30 m/s. Find the maximum height reached by the ball. Given: Initial velocity (u) = 30 m/s Final velocity at maximum height (v) = 0 m/s Acceleration due to gravity (g) = -9.8 m/s² To find: Maximum height (h) = ? Solution: Using the third equation of motion: v² = u² + 2as 0² = (30)² + 2(-9.8)h 0 = 900 - 19.6h 19.6h = 900 h = rac{900}{19.6} h ≈ 45.92 m Answer: Maximum height reached ≈ 45.92 m --- Q3. A train moving at a speed of 72 km/h applies brakes and comes to rest in 10 seconds. Calculate the retardation and distance covered during braking. Given: Initial velocity (u) = 72 km/h = 20 m/s Final velocity (v) = 0 m/s Time (t) = 10 s To find: Retardation (a) = ? Distance covered (s) = ? Solution: Using the first equation of motion: v = u + at 0 = 20 + a(10) -10a = 20 a = -2 m/s² Retardation = |a| = 2 m/s² Using the second equation of motion: s = ut + rac{1}{2}at² s = (20 × 10) + rac{1}{2}(-2)(10)² s = 200 - 100 s = 100 m Answer: Retardation = 2 m/s² Distance covered = 100 m --- Q4. A particle moves along a straight line such that its displacement from the origin is given by x(t) = t³ - 6t² + 9t + 5, where t is time in seconds. Find: a) Velocity at t = 2 s b) Acceleration at t = 3 s c) Time when velocity is zero Given: x(t) = t³ - 6t² + 9t + 5 Solution: Velocity v(t) = dx/dt = 3t² - 12t + 9 Acceleration a(t) = dv/dt = 6t - 12 a) At t = 2 s: v(2) = 3(2)² - 12(2) + 9 = 12 - 24 + 9 = -3 m/s b) At t = 3 s: a(3) = 6(3) - 12 = 18 - 12 = 6 m/s² c) When velocity is zero: 3t² - 12t + 9 = 0 t² - 4t + 3 = 0 (t - 1)(t - 3) = 0 t = 1 s or t = 3 s Answer: a) Velocity at t = 2 s = -3 m/s b) Acceleration at t = 3 s = 6 m/s² c) Time when velocity is zero = 1 s and 3 s --- Q5. Two cars start from the same point and move in opposite directions with accelerations of 2 m/s² and 3 m/s² respectively. After how much time will they be 100 meters apart? Given: Car A: u₁ = 0, a₁ = 2 m/s² Car B: u₂ = 0, a₂ = 3 m/s² Total distance between them = 100 m To find: Time (t) = ? Solution: Distance covered by Car A: s₁ = rac{1}{2}a₁t² = rac{1}{2}(2)t² = t² Distance covered by Car B: s₂ = rac{1}{2}a₂t² = rac{1}{2}(3)t² = 1.5t² Since they are moving in opposite directions: s₁ + s₂ = 100 t² + 1.5t² = 100 2.5t² = 100 t² = 40 t = √40 ≈ 6.32 s Answer: Time ≈ 6.32 seconds