A body has an initial velocity of `3 ms^-1` and has an acceleration of `1 ms^-2` normal to the direction of the initial velocity. Then its velocity `4 s` after the start is.

Options

A. `7 m s^-1` along the direction of initial velocity.
B. `7 m s^-1` along the normal to the direction of initial velocity.
C. `7 m s^-1` midway between the two directions.
D. `5 m s^-1` at an angle `tan^-1(4//3)` with the direction of initial velocity.

Answer

D

Explanation

(d) `Let u_x = 3 ms^-1, a_x = 0`
`v_y = u_y + a_y t = 0 + 1 xx 4 = 4 ms^-1`
`v = (sqrt(v_x^2 + v_y^2)) = (sqrt(3^2 + 4^2))`
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))`.