The distance travelled in n seconds =

$$S_{n} = u + \frac{1}{2} (2n – 1)a —[1]$$

So the distance travelled in t and (t + 1) seconds are

$$S_{t} = u + \frac{1}{2} (2t – 1)a —[2]$$ $$S_{t+1} = u + \frac{1}{2} (2t + 1)a —[3]$$

As given in the question:

$$S_{t} + S_{t+1} = 100 = 2 (u + at)—[4]$$

Now from the first equation of motion, the velocity of the particle after time t, if it moves with an acceleration – a is

v = u + at—[5]

Here

u = initial velocity

a = acceleration

t = time

From the equation [4] and [5], we get

V = 50cm/s

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