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

If the velocity is vecv=2hati+t^(2)hatj-9hatk then the magntidue of ac

1.	If the velocity is vecv=2hati+t^(2)hatj-9hatk then the magntidue of acceleration at t=0.5s is
Answer» `1ms^(-2)` `2ms^(-2)` ZERO `-1ms^(-2)` Solution :`a=(dv)/(DT)=(d)/(dt)(2hati+t^(2)hatj-9hatk)=2thatj` at `t=0.5simpliesa=2(0.5)=1ms^(-2)`