Question

A body has an initial velocity of `3m//s` and has an acceleration of `1m//sec^(2)` normal to the direction of the initial velocity. Then its velocity 4 seconds after the start is
A. `7m//sec` along the direction of initial velocity
B. `7m//sec` along the normal to the direction of initial velocity
C. `7m//sec` mid-way between the two directions
D. `5m//sec` at an angle off `tan^(-)(4//3)` with the direction of initial velocity.

Answer 1

Correct Answer - d
Let `u_(x)=3m//s, a_(x)=0`
`v_(y)=u_(y)+a_(y)t=0+1xx4=4ms^(-1)`
`v=sqrt(v_(x)^(2)+v_(y)^(2))=sqrt(3^(2)+4^(2))`
`=sqrt(9+16)=5m//s`
Angle made by the resultant velocity w.r.t direction of initial velocity, i.e., x-axis, is
`beta= tan^(1)((v_(y))/(v_(x))) =tan^(-1)(4/3)`