Question

A body has an initial velocity of $3m/s$ and has an acceleration of $1m{s}^{-2}$ normal to the direction of the initial velocity. Then its velocity $4s$ after the start is

• A. $7m{s}^{-1}$ along the direction of initial velocity
• B. $7m{s}^{-1}$ along the normal to the direction of the initial velocity
• C. $7m{s}^{-1}$ mid-way between the two directions
• D. $5m{s}^{-1}$ at an angle of $ta{n}^{-1}$ $\frac{4}{3}$ with the direction of the initial velocity

Answer 1

The correct option is D $5m{s}^{-1}$ at an angle of $ta{n}^{-1}$ $\frac{4}{3}$ with the direction of the initial velocity

Let $u_x=3m{s}^{-1},a_x=0$
$v_y=u_y+a_{y}t=0+1\times 4=4m{s}^{-1}$
$v=\sqrt{v_x^2+v_y^2}=\sqrt{3^2+4^2}=5m{s}^{-1}$
Angle made by the resultant velocity w.r.t. direction of initial velocity, i.e., $x-axis$, is
$\beta ={\tan }^{-1}\frac{v_y}{v_x}={\tan }^{-1}\frac{4}{3}$.