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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
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B
8 ms1
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C
5 ms1
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D
5 ms1
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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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