A kayaker spends a morning paddling on a river She travels 9

A kayaker spends a morning paddling on a river. She travels 9 miles upstream and 9 miles downstream in a total of 6 hours. In still water, she can travel at an average speed of 4 miles per hour. What is the average speed of the river\'s current in miles per hour? 1 mi/h 3 mi/h 2 mi/h 1.5 mi/h The voltage V across a capacitor at time t seconds is given by V = 146.3(1/2.72)^0.4 t Identify the voltage across the capacitor at 8 seconds to the nearest hundredth. 6.69 V 4.39 V 5.49 V 6.20 V

Solution

Let the average speed of the river’s current be x mph. Then the average speeds of the kayaker upstream and downstream are 4-x and 4 + x mph respectively. Since time = distance/speed, she travels 9 miles upstream in 9/(4-x) hours and 9 miles downstream in 9/(4+x) hours. Therefore, 9/(4-x) + 9/(4+x) = 6. Then [9(4+x) +9(4-x)]/(4-x)(4+x) = 6 or, 72/(4-x)(4+x) = 6 or, 12/(4-x)(4+x) = 1 or, (4-x)(4+x) = 12 or, 16-x² = 12 or, x² = 16-12 = 4 so that x = 2. ( x cannot be negative). Thus the average speed of the river’s current is 2 mph. The 3^rd option is the correct answer.

The voltage across a capacitor at time t seconds id given by V = 146.3(1/ 2.72)^0.41t. Thus, when t = 8, we have V = 146.3(1/ 2.72)^0.41*8 = 146.3(1/ 2.72)^3.28 = 146.3( 2.72)^-3.28 = 146.3 * 0.003755035 = 5.493616274 = 5.49 ( on rounding off to the nearest hundredth). The 3^rd option is the correct answer.