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The position vector of a particle change...

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

Text Solution

Verified by Experts

The correct Answer is:
50
`vec(r ) = 15 t^(2) hat(i) + (4 - 20 t^(2)) hat(j)`
`(vec(dt^(2)))/(dt^(2)) = 30 hat(i) + (- 40) hat(j), a = 50 m//sec^(2)`
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