Question

A body has an initial velocity of $3m{{s}^{-1}}$ 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) $7m{{s}^{-1}}$ along the direction of initial velocity
(B) $7m{{s}^{-1}}$ along the normal to the direction of initial velocity
(C) $7m{{s}^{-1}}$ mid-way between the two directions
(D) $5m{{s}^{-1}}$ at an angle of ${{\tan }^{-1}}\left( \dfrac{4}{3} \right)$with the direction of initial velocity

Answer 1

Hint: Since, the body has an acceleration in normal direction only, its velocity along the initial direction will remain absolutely constant with respect to time and the magnitude of velocity will increase along the normal direction only. We will solve them as two components of the same velocity perpendicular to each other to get the final resultant velocity.

Complete answer:
Let the direction of initial velocity be along the X-axis and the direction of perpendicular acceleration be Y-axis. Then, we have:
$\begin{align}
 & \Rightarrow {{v}_{X}}=3m{{s}^{-1}} \\
 & \Rightarrow {{a}_{X}}=0 \\
\end{align}$
And,
$\begin{align}
 & \Rightarrow {{v}_{Y}}=0 \\
 & \Rightarrow {{a}_{Y}}=1m{{s}^{-2}} \\
\end{align}$
This means the velocity of the body is constant along the X direction but increasing along the Y-direction.
Now, the magnitude of velocity along the Y-direction after 4 seconds will be equal to:
$\begin{align}
 & \Rightarrow {{v}_{Y}}=0+1(4) \\
 & \therefore {{v}_{Y}}=4m{{s}^{-1}} \\
\end{align}$
Thus, the net magnitude of velocity after 4 seconds will be equal to:
$\Rightarrow \left| \overrightarrow{v} \right|=\sqrt{{{\left( {{v}_{X}} \right)}^{2}}+{{\left( {{v}_{Y}} \right)}^{2}}}$
Putting their respective values in the above equation, we get:
$\begin{align}
 & \Rightarrow \left| \overrightarrow{v} \right|=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( 4 \right)}^{2}}} \\
 & \Rightarrow \left| \overrightarrow{v} \right|=\sqrt{9+16} \\
 & \Rightarrow \left| \overrightarrow{v} \right|=\sqrt{25}m{{s}^{-1}} \\
 & \therefore \left| \overrightarrow{v} \right|=5m{{s}^{-1}} \\
\end{align}$
Now, we will see the direction of this velocity. Let the direction of this velocity be at an angle of $\theta $ with the initial velocity vector.
Then, we can write:
$\Rightarrow \tan \theta =\dfrac{{{v}_{Y}}}{{{v}_{X}}}$
Again, putting their respective values, we get:
$\begin{align}
 & \Rightarrow \tan \theta =\dfrac{4}{3} \\
 & \therefore \theta ={{\tan }^{-1}}\left( \dfrac{4}{3} \right) \\
\end{align}$
Hence, the velocity of the body 4 seconds after the start is $5m{{s}^{-1}}$ at an angle of ${{\tan }^{-1}}\left( \dfrac{4}{3} \right)$with the direction of initial velocity.

So, the correct answer is “Option D”.

Note: The acceleration of the body in the normal direction will never affect its horizontal velocity. This is because the component of normal acceleration along the horizontal is always equal to zero. These are some basic analogies that one should be able to deduce while solving such problems or we may end up getting the wrong answer to our problems.