Question
Download Solution PDFA 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:
This question was previously asked in
UPSC CDS-I 2025 (General Studies) Official Paper (Held On: 13 Apr, 2025)
Answer (Detailed Solution Below)
Option 2 : 18 m
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Detailed Solution
Download Solution PDFThe 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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