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Sliva [168]
3 years ago
8

You want to get from a point a on the straight shore of the beach to a buoy which is 60 meters out in the water from a point b o

n the shore. b is 70 meters from you down the shore. if you can swim at a speed of 3 meters per second and run at a speed of 5 meters per second, at what point along the shore, x meters from b, should you stop running and start swimming if you want to reach the buoy in the least time possible?
Physics
1 answer:
loris [4]3 years ago
4 0
The time t for a constant speed is given by:
t = distance/ velocity

Let's call the distance to run on the beach x₁ and the remaining distance to swim x₂, then the equation becomes:
t = \frac{x_1}{5} + \frac{x_2}{3}

The distance x₂ depends on x₁:
x_2 = \sqrt{(70 - x_1)^2 + 60^2}

So: t = \frac{x_1}{5} + \frac{ \sqrt{(70 - x_1)^2 + 60^2} }{3}

Find the minimum by setting the derivative to zero: \frac{dt}{dx_1}  = 0
\frac{dt}{dx_1}  =  \frac{1}{5} -  \frac{1}{3} \frac{70 - x_1} {\sqrt{70 - x_1)^2 + 60^2} } = 0
70 - x_1 =  \frac{3}{5}   \sqrt{(70 - x_1)^2 + 60^2}  \\  \\ (70 - x_1)^2 = \frac{9}{25}((70 - x_1)^2 + 60^2) \\  \\ \frac{16}{25}(70 - x_1)^2 = \frac{9}{25}60^2 \\  \\ \frac{4}{5}(70 - x_1) = \frac{3}{5}60 \\  \\  x_1 = 25

You should stop running 70m - 25m = 45m from b.
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An airplane travels at 300 mi/h south for 2.00 h and then at 250 mi/h north for 750 miles. What is the average speed for the tri
Pie
Avg sped = total distance/ total time = 1350 mi/ 5 hrs= 270mph (i dont know if ur teacher wants you to convert this to m/s) 

300miles are traveled in 1 hr. So, 300 *2hrs = 600 miles south
750/250= 3hrs north
Total distance = 600 miles + 750 miles= 1350 miles
Total time is =  3hrs + 2hrs= 5hrs
 



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