Some additional solved numerical problems and diagrams to help you understand Rest and Motion more effectively.
Additional Solved Numerical Problems on Motion
1. Speed and Velocity Calculation
Question:
A car moves 150 meters in 10 seconds along a straight road and then returns back covering 150 meters in another 10 seconds. Find:
(a) The total distance traveled.
(b) The displacement.
(c) The speed and velocity.
Solution:
- Total Distance = 150 m (forward) + 150 m (backward) = 300 meters
- Displacement = 0 meters (Since it returns to the starting point)
- Speed = Total Distance / Total Time =30020=15 m/s= \frac{300}{20} = 15 \text{ m/s}
- Velocity = Displacement / Total Time =020=0 m/s= \frac{0}{20} = 0 \text{ m/s} Conclusion: The object has zero velocity but a non-zero speed since it returns to its starting point.
2. Acceleration and Retardation
Question:
A train starts from rest and accelerates at 2 m/s² for 10 seconds. It then moves at a constant speed for 20 seconds before decelerating at 1 m/s² until it stops. Find:
(a) Maximum speed attained.
(b) Total distance covered.
Solution:
- Maximum Speed (Using First Equation of Motion)v=u+atv = u + at v=0+(2×10)=20 m/sv = 0 + (2 \times 10) = 20 \text{ m/s}
- Distance during Acceleration (Using Second Equation of Motion)s=ut+12at2s = ut + \frac{1}{2}at^2 s=0+12×2×(10)2=100 ms = 0 + \frac{1}{2} \times 2 \times (10)^2 = 100 \text{ m}
- Distance during Uniform Motions=vt=20×20=400 ms = vt = 20 \times 20 = 400 \text{ m}
- Distance during Retardationv2=u2+2asv^2 = u^2 + 2as 0=(20)2+2(−1)×s0 = (20)^2 + 2(-1) \times s s=4002=200 ms = \frac{400}{2} = 200 \text{ m}
- Total Distance = 100 + 400 + 200 = 700 meters
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