Question

A particle moving in the x–y plane starts from the origin at \(t = 0\) with an initial velocity \((-\hat{i} + \hat{j}) \, \text{m/s}\) and undergoes an acceleration \((6\hat{i} + 4\hat{j}) \, \text{m/s}^2\). Its displacement after 2 s is:

• A. 17.32 m
• B. 14.14 m
• C. 12.42 m
• D. 10 m

Hint:

For motion in 2D, always treat displacement, velocity, and acceleration as vectors; calculate each component separately and then use Pythagoras theorem to get magnitude.

Answer 1

The Correct Option is B

Step 1: Write the position as a function of time.
With $\vec{u} = -\hat{i}+\hat{j}$ m/s and $\vec{a} = 6\hat{i}+4\hat{j}$ m/s$^2$, the displacement after time $t$ is $\vec{s} = \vec{u}t + \frac{1}{2}\vec{a}t^2$.

Step 2: Plug in $t = 2$ s component by component.
\[ s_x = (-1)(2) + \frac{1}{2}(6)(4) = -2 + 12 = 10 \]
\[ s_y = (1)(2) + \frac{1}{2}(4)(4) = 2 + 8 = 10 \]

Step 3: Spot the shortcut before reaching for a square root.
Both components come out equal, $s_x = s_y = 10$ m, so the displacement vector is simply $10(\hat{i}+\hat{j})$, pointing at $45^\circ$ to the axes. For any vector of the form $a(\hat{i}+\hat{j})$, the magnitude is just $a\sqrt{2}$.

Step 4: Compute the magnitude.
\[ |\vec{s}| = 10\sqrt{2} = 14.14\ \text{m} \]

Step 5: Conclusion.
\[ \boxed{14.14\ \text{m}} \]