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 300m exactly north of his starting point, what are the magnitude and direction of his displacement along the first straight path? 197.200 m 249 m 444 m 620 m clear all

Explanation:

Step1: Analyze x - component of displacement

Since final is north of start, $x - 260 = 0$, so $x = 260m$.

Step2: Apply Pythagorean theorem

\[s=\sqrt{260^{2}+300^{2}}\]

Step3: Calculate the value

\[s=\sqrt{67600 + 90000}=\sqrt{157600}\approx444m\]

Answer:

Let the first - stage displacement be $\vec{d}_1$, the second - stage displacement be $\vec{d}_2$ and the third - stage displacement be $\vec{d}_3$. Assume the first displacement is along the positive x - axis, the second is along the negative x - axis and the third is along the negative y - axis. Let the magnitude of the first displacement be $x$, the second be $d_2 = 260m$ and the third be $d_3=300m$.

The net displacement in the x - direction $d_x=x - 260$ and the net displacement in the y - direction $d_y=- 300$. Since the final position is directly north of the starting point, $d_x = 0$, so $x=260m$.

We want to find the magnitude of the displacement along the first straight - path.

We can use the Pythagorean theorem to find the magnitude of the net displacement from the starting point to the end - point. Since the net displacement in the x - direction is 0 (because the final position is directly north of the starting point), and the displacement in the y - direction is $d = 300m$.

The magnitude of the displacement along the first straight - path is found using the fact that the horizontal components of the displacements cancel out.

Let's consider the right - triangle formed by the displacements. The magnitude of the displacement along the first straight - path can be found using the Pythagorean theorem for the non - horizontal part of the motion.

If we consider the right - triangle with sides related to the displacements, we know that the magnitude of the displacement along the first straight - path $s$ satisfies:

\[s=\sqrt{260^{2}+300^{2}}=\sqrt{67600 + 90000}=\sqrt{157600}\approx444m\]

The direction of the first displacement: Let $\theta$ be the angle of the first displacement with the x - axis. $\tan\theta=\frac{300}{260}\approx1.154$, so $\theta=\arctan(1.154)\approx49.1^{\circ}$ below the positive x - axis.

The magnitude of the displacement along the first straight - path is approximately $444m$.