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miss Akunina [59]
3 years ago
15

Two canoeists in identical canoes exert the same effort paddling and hence maintain the same speed relative to the water. One pa

ddles directly upstream (and moves upstream), whereas the other paddles directly downstream. With downstream as the positive direction, an observer on shore determines the velocities of the two canoes to be −1.9 m/s and +2.8 m/s, respectively. What is the speed of each canoe relative to the water?
Physics
1 answer:
Serga [27]3 years ago
6 0

Answer:

Speed of river is 0.45 m/s

Speed of boat is 2.65 m/s

Explanation:

v_r = Speed of river

v_c = Speed of canoe

v_r+v_c=2.8\ m/s

v_r-v_c=-1.9\ m/s

Adding the equations we get

2v_r=0.9\\\Rightarrow v_r=\frac{0.9}{2}\\\Rightarrow v_r=0.45\ m/s

0.42+v_c=2.8\ m/s\\\Rightarrow v_c=2.8-0.45\\\Rightarrow v_c=2.65\ m/s

Speed of river is 0.45 m/s

Speed of boat is 2.65 m/s

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Answer:

solution:

to find the speed of a jogger use the following relation:  

V

=

d

x

/d

t

=

7.5

×m

i

/

h

r

...........................(

1

)  

in Above equation in x and t. Separating the variables and integrating,

∫

d

x

/7.5

×=

∫

d

t

+

C

or

−

4.7619  

=

t

+

C

Here C =constant of integration.   

x

=

0  at  t

=

0

, we get:  C

=

−

4.7619

now we have the relation to find the position and time for the jogger as:

−

4.7619  =

t

−

4.7619

.

.

.

.

.

.

.

.

.

(

2

)

Here

x  is measured in miles and  t  in hours.

(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),    

     to get:

      = −

4.7619  

      =  

1

−

4.7619

      = −

3.7619

  or  x

=

7.15

m

i

l

e

s

(b) To find the jogger's acceleration in   m

i

l

/

differentiate  

     equation (1) with respect to time.

     we have to eliminate x from the equation (1) using equation (2).  

     Eliminating x we get:

     v

=

7.5×

     Now differentiating above equation w.r.t time we get:

      a

=

d

v/

d

t

       =

−

0.675

/

      At  

      t

=

0

      the joggers acceleration is :

       a

=

−

0.675

m

i

l

/

        =

−

4.34

×

f

t

/  

(c)  required time for the jogger to run 6 miles is obtained by setting  

        x

=

6  in equation (2).  We get:

        −

4.7619

(

1

−

(

0.04

×

6  )

)^

7

/

10=

t

−

4.7619

         or

         t

=

0.832

h

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