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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

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The correct Answer is:
A
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