Question

always use significant figure rules. remember that these rules apply to all numbers that are measurements. a mountain climber hikes 25.5 km from her base camp, going 20.0° north of east. on the second day, she walks 41.0 km in a direction 65.0° north of east, at which point she has reached a summit. determine the magnitude and direction of her resultant displacement between the base camp and the second camp. (remember to analyze the two motions separately, and then add and use pythagorean theorem for the resultant and trigonometry for the angle.)

Explanation:

Step1: Resolver componentes en x - dirección

Para el primer movimiento de $d_1 = 25.5$ km con $\theta_1=20.0^{\circ}$ al norte - del - este:
$d_{1x}=d_1\cos\theta_1 = 25.5\cos(20.0^{\circ})\approx24.0$ km
Para el segundo movimiento de $d_2 = 41.0$ km con $\theta_2 = 65.0^{\circ}$ al norte - del - este:
$d_{2x}=d_2\cos\theta_2=41.0\cos(65.0^{\circ})\approx17.3$ km
La componente total en x es $d_x=d_{1x}+d_{2x}=24.0 + 17.3=41.3$ km

Step2: Resolver componentes en y - dirección

Para el primer movimiento: $d_{1y}=d_1\sin\theta_1=25.5\sin(20.0^{\circ})\approx8.72$ km
Para el segundo movimiento: $d_{2y}=d_2\sin\theta_2=41.0\sin(65.0^{\circ})\approx37.1$ km
La componente total en y es $d_y=d_{1y}+d_{2y}=8.72 + 37.1 = 45.8$ km

Step3: Calcular la magnitud del desplazamiento resultante

Usando el teorema de Pitágoras $d=\sqrt{d_x^{2}+d_y^{2}}$
$d=\sqrt{(41.3)^{2}+(45.8)^{2}}=\sqrt{1705.69 + 2097.64}=\sqrt{3803.33}\approx61.7$ km

Step4: Calcular la dirección del desplazamiento resultante

Usando trigonometría, $\theta=\tan^{- 1}(\frac{d_y}{d_x})$
$\theta=\tan^{-1}(\frac{45.8}{41.3})\approx48.0^{\circ}$ al norte - del - este

Answer:

Magnitud: $61.7$ km
Dirección: $48.0^{\circ}$ al norte - del - este