Question

1. the riverboat crosses an 80m wide river with a boat velocity of 5 m/s east and a river current of 4 m/s south.

a. what is the velocity of the riverboat as measured by a person standing next to the river?
b. how long does it take the riverboat to cross the river?
c. how far down - river will the boat move while it crosses?
d. how will the time to cross the river and the distance the boat travel downriver while crossing change if:
i. the river current has slowed to 2 m/s?
ii. the river current has slowed to 2 m/s and the riverboat travels at 10 m/s?

Explanation:

Step1: Find the resultant velocity in part a

The boat has a velocity of $v_x = 5$ m/s (East - across the river) and $v_y= 4$ m/s (South - due to river current). Using the Pythagorean theorem for vector addition, the resultant velocity $v=\sqrt{v_x^{2}+v_y^{2}}$. So, $v = \sqrt{5^{2}+4^{2}}=\sqrt{25 + 16}=\sqrt{41}\approx 6.4$ m/s.

Step2: Calculate time to cross the river in part b

The width of the river $d = 80$ m and the velocity across the river $v_x=5$ m/s. Using the formula $t=\frac{d}{v_x}$, we have $t=\frac{80}{5}=16$ s.

Step3: Calculate downstream - distance in part c

The time to cross the river $t = 16$ s and the velocity of the river - current $v_y = 4$ m/s. Using the formula $x=v_y\times t$, we get $x = 4\times16=64$ m.

Step4: Analyze changes in part d - i

The time to cross the river $t=\frac{d}{v_x}$, where $d$ is the width of the river and $v_x$ is the velocity of the boat across the river. Since $d$ and $v_x$ are unchanged, the time to cross the river remains the same, $t = 16$ s. The downstream distance $x=v_y\times t$. With $v_y$ changing from 4 m/s to 2 m/s and $t = 16$ s, the new downstream distance $x'=2\times16 = 32$ m. So the time to cross the river stays the same and the downstream distance decreases.

Step5: Analyze changes in part d - ii

The time to cross the river $t=\frac{d}{v_x}$, with $d = 80$ m and $v_x = 10$ m/s, so $t=\frac{80}{10}=8$ s. The downstream distance $x=v_y\times t$, with $v_y = 2$ m/s and $t = 8$ s, so $x=2\times8 = 16$ m. The time to cross the river decreases and the downstream distance decreases.

Answer:

a. Approximately 6.4 m/s
b. 16 s
c. 64 m
d. i. Time to cross stays the same (16 s), downstream distance decreases to 32 m.
ii. Time to cross decreases to 8 s, downstream distance decreases to 16 m.