1.

Velocity of a particle at time `t=0` is `2ms^(-1)`. A constant acceleration of `2ms^(-2)` acts on the particle for `1 second` at an angle of `60^(@)` with its initial velocity. Find the magnitude of velocity at the end of `1 second`.A. `sqrt3 m//s`B. `2sqrt3 m//s`C. `4 m//s`D. `8 m//s`

Answer» Correct Answer - B
`vecv=v_(x)hati+v_(y)hatj:v_(x)=u_(x)+a_(x)t,v_(y)=u_(y)+a_(y)t`
`a_(x)=a cos theta,a_(y)=a cos theta`


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