A motorcycle has initial velocity of 5 m/s. After 3 seconds, the velocity is 7 m/s. The displacement of the motorcycle in 3 seconds is:

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UPSC CDS-I 2025 (General Studies) Official Paper (Held On: 13 Apr, 2025)
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  1. 21 m
  2. 18 m
  3. 36 m
  4. 6 m

Answer (Detailed Solution Below)

Option 2 : 18 m
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The correct answer is Option 2 (18 m).

Key Points

  • The displacement of the motorcycle is calculated using the formula for uniformly accelerated motion: s = (u + v)/2 × t, where 'u' is the initial velocity, 'v' is the final velocity, and 't' is the time.
  • Given values: Initial velocity (u) = 5 m/s, Final velocity (v) = 7 m/s, Time (t) = 3 seconds.
  • Substituting the values: s = (5 + 7)/2 × 3 = 12/2 × 3 = 18 m.
  • The motorcycle travels a displacement of 18 meters in 3 seconds under uniform acceleration.
  • This result is consistent with the kinematic equations of motion for uniformly accelerated systems.

Additional Information

  • Uniformly Accelerated Motion:
    • It refers to motion where the velocity of an object changes by an equal amount in every equal time interval.
    • The kinematic equations of motion are applicable for this type of motion, which include formulas for displacement, velocity, and acceleration.
  • Kinematic Equation for Displacement:
    • The formula s = (u + v)/2 × t is used to calculate displacement when initial velocity, final velocity, and time are known.
    • This equation is derived based on the assumption of uniform acceleration.
  • Relation Between Velocity and Acceleration:
    • Acceleration is the rate of change of velocity with respect to time and is constant in uniformly accelerated motion.
    • It can be calculated using the formula: a = (v - u)/t.
  • Displacement vs. Distance:
    • Displacement is a vector quantity representing the shortest path between the initial and final points.
    • Distance is a scalar quantity representing the total length of the path traveled, irrespective of direction.
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