Question

# A body of mass $$3\ kg$$ moving with a constant acceleration covers a distance of $$10\ m$$ in the $$3^{rd}$$ second and $$Ibm$$ in the $$4^{th}$$ second respectively. The initial velocity of the body is:

A
10 ms1
B
8 ms1
C
5 ms1
D
5 ms1

Solution

## The correct option is D $$-5\ ms^{-1}$$Given that, Mass of body = $$3\ kg$$ Distance covered in nth second = $$u+a/2 (2n-1)$$  initial velocity = $$u$$  nth second = $$n$$  acceleration = $$a$$ Now, distance covered in 3rd second =$$10\ m$$   $$10=u+\dfrac{a}{2}\left( 2\times 3-1 \right)$$  $$20=2u+5a.... (I)$$ Now, distance covered in 4rd second= 16 m   $$16=u+\dfrac{a}{2}\left( 2\times 4-1 \right)$$  $$32=2u+7a... (II)$$ Now, from equations (I) and (II)   $$-12=-2a$$  $$a=6\,m/{{s}^{2}}$$ Now, put the value of a in equation (I)   $$20=2u+5a$$  $$2u=20-30$$  $$u=-5\,m/s$$ Hence, the initial velocity is $$-5\ m/s$$  Physics

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