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The position vector of a particle change...

The position vector of a particle changes with time according to the relation `vec(r ) (t) = 15 t^(2) hat(i) + (4 - 20 t^(2)) hat(j)` What is the magnitude of the acceleration at t = 1?

A

25

B

100

C

40

D

50

Text Solution

Verified by Experts

The correct Answer is:
D
`vec(a) = (d^(2)vec(r))/(dt^(2))`
`vec (a) = 30 hat(i) - 40 hat(j)`
`rArr a_("net")= sqrt(30^(2)+40^(2))rArr|a_("net")|=50m//s^(2)`
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