Help needed with a problem

[murshid]

Here is the problem:

Quote

"A shark, looking for dinner, is swimming parallel to a straight beach and 90 feet offshore. The shark is swimming at the constant speed of 30 feet per second. At time t=0, the shark is directly opposite a lifeguard station. How fast is the shark moving away from the lifeguard station when the distance between them is 150 feet?"

Now, if the speed is constant, then shouldn't the answer be 30 feet per second no matter where the shark is?

[Ghideon]

11 minutes ago, murshid said:

Now, if the speed is constant, then shouldn't the answer be 30 feet per second no matter where the shark is?

Hint 1: The shark moves parallel to the straight beach. Compare distance from lifeguard at time t=-0.5s and time t=0.5s. Does that validate or invalidate your proposal?
Hint 2: Draw a picture of the situation and see if you proposal applies.

[Strange]

11 minutes ago, murshid said:

Here is the problem:

Now, if the speed is constant, then shouldn't the answer be 30 feet per second no matter where the shark is?

There is a difference between the speed that shark is going (e.g. the speed it is going past the lifeguard station) and the speed it is moving away from the lifeguard station.

To take the extreme case, at the instant that the shark is just passing the lifeguard station, it is not moving away at all: its distance from the lifeguard station is pretty much constant (90 feet). 

Perhaps you need to start by drawing a diagram with lines showing the distance between the lifeguard station and the shark at fixe intervals. That might help you visualise what is going on.

1 minute ago, Ghideon said:

Hint 2: Draw a picture of the situation and see if you proposal applies.

Sound like that is a popular suggestion. (It was pretty much the first thing we were told to do for every problem in maths)

[murshid]

25 minutes ago, Strange said:

There is a difference between the speed that shark is going (e.g. the speed it is going past the lifeguard station) and the speed it is moving away from the lifeguard station.

That's the part I did not understand. I did draw a diagram, but got confused by the language. So, I attached the diagram. 

O is the point where the lifeguard is.
A is the point where the shark is at t = 0. So, OA = 90 feet. 
B is the point where the shark is 150 feet away from the lifeguard. So, OB = 150 feet. 

AB = sqrt (OB² - OA²) = sqrt (150² - 90²) = 120 feet

Since the shark is moving at a constant speed of 30 ft./s, it would reach B at t = 4 ; [ t = d/v =120/30 = 4 ]

So, in 4 seconds, the shark went 150-90 = 60 feet further away from the lifeguard. 

The speed at which the shark is moving away from the lifeguard at B = 60/4 = 15 ft./s

Is that it?

[Strange]

1 hour ago, murshid said:

So, in 4 seconds, the shark went 150-90 = 60 feet further away from the lifeguard. 

The speed at which the shark is moving away from the lifeguard at B = 60/4 = 15 ft./s

Is that it?

Close. That is the average speed between t=0 and t=4 (and as you can see, it is less than the 30 ft/s of the sharks speed).

I'm afraid it is really late here and I am too tired to suggest a good way to work out the speed at t=4! But you might want to think about the fact it starts at 0, gets up to v (the speed we want to find) and the average is 15. 

[Country Boy]

Is this a "Calculus" problem?  The "standard" way to calculate (or even define) "speed at an instant" is the derivative with respect to time.  The shark is a constant 90 feet from the shore and moving along the shore at a constant 30 feet per second so its distance down the shore from the lifeguard station is 30t feet where t is the time, in seconds, since the shark passed the lifeguard station.  By the Pythagorean theorem, the straight line distance from the lifeguard station to the shark, at time "t", is given by [math]D^2= 90^2+ (30t)^2[math].  Differentiating both sides, [math]2D (dD/dt)= 2(30)(30t)= 1800t[/math]. The problem asked for the distance when the shark is D= 150 feet past the lifeguard station but we need to also calculate t at that point.  Use [math]D^2= 150^2= 22500= 90^2+ (30t)^2= 8100+ 900t^2[/math].  So we need to solve [math]900t^2= 22500- 8100= 14400[/math] or [math]t^2= 14400/900= 16[/math].  t= 4 and D= 150 so $2D(dD/dt)= 1800t$ becomes $300 (dD/dt)= 7200$.  The rate at which the straight line distance from the lifeguard station to the shark is increasing, when the shark is 150 feet from the lifeguard station, is [math]dD/dt= 7200/300= 24[/math] feet per second.

Edited April 21, 2020 by Country Boy

[joigus]

1) "How fast the shark is moving away from the lifeguard station" requires you to think about vectors. Picture an imaginary straight line lifeguard-shark and try to think how it changes.

2) The datum is the speed or rapidity (the norm, or intensity, or "modulus" of the vector), not that velocity. That's another line (parallel to the coast.)

3) Think Pythagoras. He was a very wise guy, or maybe a bunch of guys, nobody knows to this day.

And I can't give you any more clues.

Edited May 3, 2020 by joigus
mistyped