A particle moves along a straight line such that its acceleration is given by a = -k v^2, where k is a positive constant and v is velocity. If initial velocity at t = 0 is v0, what is the distance covered by the particle when its velocity drops to v0 / 2?
Answer and explanation
Correct answer: C. (ln 2) / k
- Step 1: Acceleration a = v (dv/ds) = -k v^2.
- Step 2: dv/v = -k ds.
- Step 3: Integrate from v0 to v0/2: ∫_(v0 to v0/2) dv/v = -k ∫_(0 to s) ds.
- Step 4: ln((v0/2) / v0) = -k s => ln(1/2) = -k s => -ln(2) = -k s => s = (ln 2) / k.
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