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BITSAT 2011 Paper

Section: Physics

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Question : 2 of 150

Marks: +1, -0

A particle is moving in a straight line with initial velocity and uniform acceleration a. If the sum of the distance travelled in t^th and (t + 1)^th seconds is 100 cm, then its velocity after

t

seconds, in cm/s, is

80
50
20
30

Solution:

The distance travelled in

n^{\;\text{th }\;}

second is

S_n = u + \;\frac{1}{2}(2n - 1)a

So distance travelled in

t^{th} \& (t+1)^{\;\text{th }\;}

second are

S_t = u + \;\frac{1}{2}(2t - 1)a

S_{t+1} = u + \;\frac{1}{2}(2t + 1)a

As per question,

S_t + S_{t+1} = 100 = 2(u + at) \;\;\;\;\; \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (1)

Now from first equation of motion the velocity of particle after time

t

, if it moves with an acceleration a is

v = u + at \;\;\;\;\; \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)

where

u

is initial velocity So from equation

(1)

and

(2)

, we get

v = 50\,\text{cm/s}

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