A particle has initial velocity $(3\hat{i}+4\hat{j})$ and has acceleration $(0.4\hat{i}+0.3\hat{j})$. Its spee

Question

A particle has initial velocity $(3\hat{i}+4\hat{j})$ and has acceleration $(0.4\hat{i}+0.3\hat{j})$. Its speed after $10 	ext{ s}$ is:

Accepted Answer

1. **Identify Given Values:** 
   - Initial velocity $\vec{u} = 3\hat{i} + 4\hat{j}$
   - Acceleration $\vec{a} = 0.4\hat{i} + 0.3\hat{j}$
   - Time $t = 10 	ext{ s}$
2. **Apply Kinematic Equation:** For motion with constant acceleration in a plane, the velocity vector at time $t$ is given by $\vec{v} = \vec{u} + \vec{a}t$ .
3. **Calculate Final Velocity Vector:**
   $$ \vec{v} = (3\hat{i} + 4\hat{j}) + 10(0.4\hat{i} + 0.3\hat{j}) $$
   $$ \vec{v} = (3\hat{i} + 4\hat{j}) + (4\hat{i} + 3\hat{j}) $$
   $$ \vec{v} = (3+4)\hat{i} + (4+3)\hat{j} = 7\hat{i} + 7\hat{j} $$
4. **Calculate Speed:** Speed is the magnitude of the velocity vector.
   $$ |\vec{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{7^2 + 7^2} = \sqrt{49 + 49} = \sqrt{98} = 7\sqrt{2} 	ext{ units} $$

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