Question

A particle moving with uniform acceleration has velocity of 6 ms-1 at a distance 5 m from the initial position. After moving another 7 m the velocity becomes 8 ms-1. The initial velocity and acceleration of the particle are 1) 2 ms-l; 4 ms-1 2) 4 ms-1 ; 2 ms-2 3) 4 ms-1; 4 ms-2 4) 6 ms-1 ; 1 ms-2 ​

Answer 1

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Let acceleration be a m/s^2 and initial velocity be u m/s.

Now s(distance) in first part is 5 m and final velocity is 6m/s

therefore, using- v^2-u^2=2as we get,

6^2-u^2=2*a*5

=> 36-u^2=10a   --------- eq.1

now in second part of journey...6m/s becomes initial velocity, 8m/s becomes final velocity and 7m is s

again applying the same equation...we get,

8^2 - 6^2=2*a*7

=> 64-36=14a

=>28=14a

=>a=2m/s^2

NOW SUBSTITUTING VALUE OF a IN EQ. 1, WE GET-

36-u^2=10*2

=>36-u^2=20

=>36-20=u^2

=>16=u^2

=>u=4m/s

Therefore, we get- acceleration = 2m/s^2 and initial velocity= 4m/s