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If the velocity is vecv=2hati+t^(2)hatj-...

If the velocity is `vecv=2hati+t^(2)hatj-9hatk` then the magntidue of acceleration at `t=0.5s` is

A

`1ms^(-2)`

B

`2ms^(-2)`

C

zero

D

`-1ms^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`a=(dv)/(dt)=(d)/(dt)(2hati+t^(2)hatj-9hatk)=2thatj`
at `t=0.5simpliesa=2(0.5)=1ms^(-2)`
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