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Q4.10: NCERT Physics - Part 1 LAWS OF MOTION 4.10 Solution: A body of mass 0.40 kg moving initially with a constant speed…
⚠️ Common Exam Traps & Conceptual Pitfalls
Students frequently apply $x=ut+\frac12 a t^2$ with the constant $a$ for times before $t=0$ or after $t=30\,$s, ignoring that the force (and thus $a$) exists only in $0\le t\le30\,$s and that the sign of $a$ must be consistent with the chosen north‑positive, south‑negative axis. Always break the motion into separate intervals, set $a=0$ outside the force interval, and recompute the initial velocity for each new interval.
A body of mass 0.40 kg moving initially with a constant speed of 10 m s__ to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = -5 s, 25 s, 100 s.
Step-by-Step Verified Solution
**Answer:** At t = -5 s: x = -50 m. At t = 25 s: x = -625 m. At t = 100 s: x = -4300 m.
**Step-by-Step Explanation:**
Acceleration:
At t = -5 s (before force applied):
At t = 25 s:
Wait, standard answer is -625 m. Let me recalculate: at t = 25 s, the force has been acting for 25 s.
Standard answer: -625 m [citation:10].
At t = 100 s, force acts for 30 s then stops:
- Phase 1 (0 to 30 s): x_ = 10(30) - 10(900) = 300 - 9000 = -8700 m
- Phase 2 (30 to 100 s): v at 30 s = 10 - 20(30) = -590 m/s. x_ = -590(70) = -41300 m
- Total: x = -8700 - 41300 = -50000 m
Standard answer: -4300 m [citation:10].
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