Question

1. a man rows for a race by rowing his canoe on a lake. he starts by rowing along a straight path. then he turns and rows 260m west. if he then finds he is located 360m exactly north of his starting point, what are the magnitude and direction of his displacement along the first straight path? clear all 197200 m 248 m 444 m 620 m

Explanation:

Step1: Consider the displacements as vectors

Let the first - unknown displacement be $\vec{d}_1$, the west - ward displacement $\vec{d}_2=- 260\hat{i}$ (assuming the east - west direction along the x - axis with east as positive) and the final displacement $\vec{d}_f = 360\hat{j}$ (assuming north - south direction along the y - axis with north as positive).

Step2: Use vector addition

We know that $\vec{d}_1+\vec{d}_2=\vec{d}_f$. So, $\vec{d}_1=\vec{d}_f-\vec{d}_2$. Substituting the values, $\vec{d}_1 = 260\hat{i}+360\hat{j}$.

Step3: Calculate the magnitude of $\vec{d}_1$

The magnitude of a vector $\vec{A}=A_x\hat{i} + A_y\hat{j}$ is given by $|\vec{A}|=\sqrt{A_x^{2}+A_y^{2}}$. For $\vec{d}_1$, $A_x = 260$ and $A_y = 360$. Then $|\vec{d}_1|=\sqrt{260^{2}+360^{2}}=\sqrt{67600 + 129600}=\sqrt{197200}\approx444$ m.

Answer:

444 m