# 1. During one part of its migration, a salmon is swim upstream at 6 mph. The current flows downstream at 3 mph at an angle of 7°. How fast is the salmon going upstream? 2. Juan sits on sled on...

1. During one part of its migration, a salmon is swim upstream at 6 mph. The current flows downstream at 3 mph at an angle of 7°. How fast is the salmon going upstream?

2. Juan sits on sled on side of hill inclined 45°. Both weights of Juan and sled is 140 pounds. What force is requires to keep sled from sliding down the hill?

3. Find the work done by 10 pound force acting in direction (1, 2) moving object 3 feet from (0, 0) to (3, 0).

4. For nondegenerate quadratic equation, explain what the following tell you about the graph of the equation.

a) B^2-4AC=0

b) B^2-4AC>0

c) B^2-4AC<0

d)B not equal to 0.

5. Quadratic equation. Explain how the graph of this equation changes as B goes from 0 to 5.

6. Jose and Aparna were attempting to rotate the axes to eliminate the xy term for the equation . Jose used a=-`pi` /8. Aparna used a=3`pi` /8.

The students got different equations on x and y values, however both are correct. Why is it possible?

llltkl | College Teacher | (Level 3) Valedictorian

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1. The Salmon swims upstream with a velocity of 6 mph at an angle of 7`^o` with respect to the river stream, which has a velocity of 3 mph. The situation is represented in the attached figure (i).

From figure (i), we have

`stackrel (rarr)(AC)=lt3cos(-83^o), 3sin(-83^o)gt`

`=lt0.37, -2.98gt`

`stackrel (rarr)(BA)=lt0,6gt`

Therefore,

`stackrel (rarr)(BC)=stackrel (rarr)(BA)+stackrel (rarr)(AC)`

`=lt0.37,(-2.98+6)gt`

`=lt0.37, 3.02gt`

So, `|stackrel (rarr)(BC)|=sqrt((0.37)^2+(3.02)^2)`

`=3.04 mph` (approx.)

Therefore, absolute velocity of the salmon = 3.04 mph upstream.

2. Force due to gravity (weight) acts vertically downwards and is given by,

mg =140 lbs
It can be resolved into two mutually perpendicular components as shown in the attached figure (ii).

The component of weight vector responsible for the downward sliding motion of the sled is

`mgsin45^o` .

In order to keep the sled from falling, this downward force has to be balanced by two forces,

1. Frictional force

2. Upward pulling force.

Assuming the surface to be frictionless, the upward pulling force required to keep the sled from falling is `=mgsin45^o`

`=140 sin45^o`

`= 99 lbs` (approx.)

Therefore, an upward pulling force of 99 lbs is required to keep Juan from sliding down.

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