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Gaurav |spideyunlimited| Ragtah (4463)

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3. A greatest acceleration and retardation that a car may have is 'a'. The minimum time in which car can cover 'd' distance.

Let it cover distance 'x' with acceleration a, and 'd-x' with retardation a.

x = 1/2 * a (t1)^2
t1 = root (2x/a)
v = root(2ax)

For second half of motion
d - x = root(2ax).t2 - 1/2*a.(t2)^2
t2 = root(2x/a) +- root [2(2x-d)/a]

So Total Time = t1 + t2
= root(2x/a) + root(2x/a) +- root [2(2x-d)/a]
As we're finding the minimum time, we only consider -->
root(2x/a) + root(2x/a) - root [2(2x-d)/a]
Now when u minimize this function, you get x = d/2
So the time comes out as:
T = 2.root(d/a)

- Gaurav Ragtah (spideyunlimited)

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