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:

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.