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Question: A particle is moving in x-y-plane at 2 m/s along x-axis, 2 seconds later, its velocity is 4 m/s in a...

A particle is moving in x-y-plane at 2 m/s along x-axis, 2 seconds later, its velocity is 4 m/s in a direction making 60° with positive x-axis. Its average acceleration for this period of motion is:

A

√5 m/s², along y-axis

B

√3 m/s², along y-axis

C

√5 m/s², along at 60° with positive x-axis

D

3m/s², at 60° with positive x-axis.

Answer

√3 m/s², along y-axis

Explanation

Solution

  • Initial velocity: The particle moves along the x-axis at 2 m/s. v1=2i^\vec{v_1} = 2 \hat{i} m/s

  • Final velocity: After 2 seconds, its velocity is 4 m/s in a direction making 60° with the positive x-axis. v2=(4cos60)i^+(4sin60)j^\vec{v_2} = (4 \cos 60^\circ) \hat{i} + (4 \sin 60^\circ) \hat{j} v2=(4×12)i^+(4×32)j^\vec{v_2} = (4 \times \frac{1}{2}) \hat{i} + (4 \times \frac{\sqrt{3}}{2}) \hat{j} v2=2i^+23j^\vec{v_2} = 2 \hat{i} + 2\sqrt{3} \hat{j} m/s

  • Time interval: Δt=2\Delta t = 2 s

  • Average acceleration: aavg=ΔvΔt=v2v1Δt\vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t} = \frac{\vec{v_2} - \vec{v_1}}{\Delta t} aavg=(2i^+23j^)(2i^)2\vec{a}_{avg} = \frac{(2 \hat{i} + 2\sqrt{3} \hat{j}) - (2 \hat{i})}{2} aavg=23j^2\vec{a}_{avg} = \frac{2\sqrt{3} \hat{j}}{2} aavg=3j^\vec{a}_{avg} = \sqrt{3} \hat{j} m/s²

  • Magnitude: aavg=3|\vec{a}_{avg}| = \sqrt{3} m/s²

  • Direction: Along the y-axis.