A particle moves with constant acceleration. Its positions at times 0, T and 2T are respectively 0, D and -D, where D>0. The total distance travelled from t=0 to t=2T is:
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Question

A particle moves with constant acceleration. Its positions at times 0, T and 2T are respectively 0, D and −D, where D>0. The total distance travelled from t=0 to t=2T is:

G1 Physics · Accepted Answer

Let x = ut + (a/2)t². At t=T: uT + aT²/2 = D. (1) At t=2T: 2uT + 2aT² = −D. (2) Let A=uT, B=aT²/2. Then A+B=D, 2A+4B=−D. Solving, B=−3D/2, A=5D/2. Thus u = 5D/(2T), a = −3D/T². Turning time: t₀ = −u/a = 5T/6. Position there: x₀ = ut₀ + (a/2)t₀² = 25D/24. The particle moves from 0 to +25D/24, then reverses and ends at −D. Total distance = 25D/24 + [25D/24 + D] = 50D/24 + 24D/24 = 74D/24 = 37D/12. Therefore option B is correct.