t=x2+xdxdt=dxd[x2+x]v1=2x+1 v=(2x+1)−1...(i) aaa=dxvdv=(2x+1)−1[−1(2x+1)−2×2]=−1(2x+1)−3×2=−2(2x+1)−3=−(2x+1)32
NEET UG 2025 — Physics Mechanics
In some appropriate units, time (t) and position ( x ) relation of a moving particle is given by t=x2+x. The acceleration of the particle is
Held on 30 Apr 2025 · Verified 9 Jul 2026.
−(x+2)32
−(2x+1)32
+(x+1)32
+2x+12
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A bullet of mass 10 g moving with 400 m/s penetrates a wall and comes to rest. The loss in kinetic energy is:
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