Question

Initial velocity vector vA has a magnitude of 3.00 meters per second and points 20.0o north of east, while final velocity vector vB has a magnitude of 6.00 meters per second and points 40.0o south of east. Find the magnitude and the direction of the change in velocity vector Δv (which is the vector subtraction of the two vectors: final velocity vector minus initial velocity vector).

Answer 1

Answer:

$5.2\ \text{m/s}$

$70^{\circ}$ south of east

Explanation:

$v_a$ = 3 m/s

$\theta_a$ = $20^{\circ}$ north of east

$v_b$ = 6 m/s

$\theta_b$ = $40^{\circ}$ south of east = $360-40=320^{\circ}$ north of east

x and y component of $v_a$

$v_{ax}=v_a\cos \theta\\\Rightarrow v_{ax}=3\times \cos 20^{\circ}\\\Rightarrow v_{ax}=2.82\ \text{m/s}$

$v_{ay}=v_a\sin\theta\\\Rightarrow v_{ay}=3\times \sin20^{\circ}\\\Rightarrow v_{ay}=1.03\ \text{m/s}$

x and y component of $v_b$

$v_{bx}=v_b\cos \theta\\\Rightarrow v_{bx}=6\times \cos 320^{\circ}\\\Rightarrow v_{bx}=4.6\ \text{m/s}$

$v_{by}=v_b\sin\theta\\\Rightarrow v_{by}=6\times \sin320^{\circ}\\\Rightarrow v_{by}=-3.86\ \text{m/s}$

$\Delta v=v_b-v_a\\\Rightarrow \Delta v=(4.6-2.82)\hat{i}+(-3.86-1.03)\hat{j}\\\Rightarrow \Delta v=1.78\hat[i}-4.89\hat{j}$

Magnitude

$|\Delta v|=\sqrt{(-4.89)^2+1.78^2}\\\Rightarrow \Delta v=5.2\ \text{m/s}$

Direction

$\theta=\tan{-1}|\dfrac{-4.89}{1.78}|\\\Rightarrow \theta=70^{\circ}$

The magnitude of the change in velocity vector is $5.2\ \text{m/s}$ and the direction is $70^{\circ}$ south of east.