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A particle starting from rest accelerates uniformly and travels a distance $$d$$ for some time. It then travels uniformly at the speed acquired after the initial acceleration for an equal time, after which it decelerates and comes to rest. The magnitude of the deceleration is twice the initial acceleration. If total distance travelled is $$nd$$, find $$n$$.
Correct Answer: 3.5
With acceleration $$a$$ for time $$t$$, $$d = \frac{1}{2}at^2$$ and the speed reached is $$at$$. At constant speed for time $$t$$ it covers $$at^2 = 2d$$. Decelerating at $$2a$$ it covers $$\frac{(at)^2}{2 \times 2a} = \frac{at^2}{4} = \frac{d}{2}$$. The total is $$d + 2d + \frac{d}{2} = 3.5d$$, so $$n = 3.5$$.
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