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    Question

    A particle starts from rest and accelerates uniformly

    at 2 m/s² for 10 seconds, then moves with constant velocity for 5 seconds, and finally decelerates at 4 m/s² until it stops. What is the total distance travelled?
    A 200 m Correct Answer Incorrect Answer
    B 275 m Correct Answer Incorrect Answer
    C 225 m Correct Answer Incorrect Answer
    D 250 m Correct Answer Incorrect Answer

    Solution

    We divide the motion into three parts and calculate the distance for each: Part 1: Uniform acceleration Initial velocity = 0, acceleration = 2 m/s², time = 10 s Distance:
    s₁ = (1/2) × a × t² = (1/2) × 2 × (10)² = 100 m Final velocity after acceleration: v = u + at = 0 + 2×10 = 20 m/s   Part 2: Constant velocity motion Velocity = 20 m/s, time = 5 s Distance: s₂ = v × t = 20 × 5 = 100 m   Part 3: Uniform deceleration to rest Initial velocity = 20 m/s, deceleration = 4 m/s² Using v² = u² + 2as: 0 = (20)² + 2×(–4)×s₃ ⇒ 400 = 8s₃ ⇒ s₃ = 50 m Total distance = s₁ + s₂ + s₃ = 100 + 100 + 50 = 250 m

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