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Q. A particle moving with uniform acceleration has velocity of $6\, m \,s ^{-1}$ at a distance 5 m from the initial position. After moving another 7 m the velocity becomes $8 \,m \,s ^{-1}$. The initial velocity and acceleration of the particle are

Motion in a Straight Line

Solution:

Using, $v^{2}=u^{2}+2 a s$,
For second part of motion, we have
$(8)^{2}=(6)^{2}+2 \times a \times 7$ or $a=2 m s ^{-2}$
For first part of motion,
$(6)^{2}=u^{2}+2 \times 2 \times 5$ or $u=4 m s ^{-1}$