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, $4$ seconds after the start is

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

Answer 1

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

In $x$ direction, $V_x=3$m/s
In $y$ direction, $V_y=at=1\left(4\right)=4$m/s.
$V_{net}=\sqrt{V_x^2+V_y^2}=\sqrt{3^2+4^2}=5$m/s

at an angle ${\tan }^{-1}\left(4/3\right)$ with initial direction of velocity.