Q 11-03-014JEE MainAIEEE 2009Easy
A particle has an initial velocity $3\hat{i} + 4\hat{j}$ and an acceleration of $0.4\hat{i} + 0.3\hat{j}$. Its speed after $10$ s is
Answer: (A) $7\sqrt{2}$ units
Concept. With constant acceleration, $\vec v=\vec u+\vec at$ holds as a vector equation, so each component changes independently. The speed is the magnitude of the final velocity vector.
Step 1. Velocity after $10$ s.
$$\vec v=(3\hat i+4\hat j)+(0.4\hat i+0.3\hat j)(10)=(3+4)\hat i+(4+3)\hat j=7\hat i+7\hat j$$
Step 2. Speed.
$$|\vec v|=\sqrt{7^2+7^2}=7\sqrt2\ \text{units}$$
Answer. $7\sqrt2$ units (Option A). Trap: the speed is not $7+7=14$; add the components as a vector.
Solution by Sreeraj P, M.Sc Physics
