A body moving with uniform acceleration in a straight line is at points A, B, C, D after successive equal intervals of time. The distance AD is equal to:

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2(BC)

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3(BC)

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4(BC)

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Solution

The correct option is D 3(BC)
Let the starting point is O. And the body reach,
A at time t,
B at time 2t.
C at time 3t,
D at time 4t.
Using $s = u t + \frac{1}{2} a t^{2}$
O to A is $u t + \frac{t^{2}}{2}$
O to B is $u t + \frac{4 t^{2}}{2}$
O to C is $u t + \frac{9 t^{2}}{2}$
O to D is $u t + \frac{16 t^{2}}{2}$
So AD is (OD-OA)= $\frac{16 t^{2}}{2} + u t$ $- \frac{t^{2}}{2} - u t$ $= \frac{15 t^{2}}{2}$
And BC is (OC-OB) = $\frac{9 t^{2}}{2} + u t$ $- \frac{4 t^{2}}{2} - u t$ $= \frac{5 t^{2}}{2}$
Therefore, AD=3BC
Hence option C is correct.

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