Q.

At time t = 0 a particle starts travelling from a height 7zcm in a plane keeping z coordinate constant. At any instant time it's position along the x and y directions are defined as 3t and 5t3 respectively. At t = 1s acceleration of the particle will be

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a

-30y^

b

3x^+15y^+7z

c

30y^

d

3x^ + 15y^

answer is B.

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Detailed Solution

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r = 3ti^ + 5t3j^ + 7k^

d2rdt2 = 30tj^ At t =1 

d2rdt2 = 30j^

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