Question

1. takashi trains 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?

Explanation:

Step1: Identify vector components

Let the first - displacement be along the first straight - path (unknown length, say \(x\)), the second displacement \(A = 260m\) west (negative \(x\) - direction) and the resultant displacement \(R = 360m\) north (positive \(y\) - direction).

Step2: Apply the Pythagorean theorem

We know that, if we consider the right - triangle formed by the vectors, according to the Pythagorean theorem \(R^{2}=x^{2}+A^{2}\). We want to find \(x\). Rearranging the formula for \(x\), we get \(x=\sqrt{R^{2}-A^{2}}\).
Substitute \(R = 360m\) and \(A = 260m\) into the formula:
\[x=\sqrt{360^{2}-260^{2}}=\sqrt{(360 + 260)(360 - 260)}=\sqrt{620\times100}=\sqrt{62000}\approx 248m\]

Answer:

248 m