1. Assume the velocity of the Ohio River at Cincinnati on October 21st is 0.900m

Question

1. Assume the velocity of the Ohio River at Cincinnati on October 21st is 0.900m/s to the West. If you are in a kayak you can row at a constant rate of 2.4 m/s in still water. The Ohio River is 800. m wide.

i have already found that it took me 333 sec to cross the river, pointed perpedicularly, and i was 300 m downriver after crossing...

i now need to find:

a. If you now paddle the kayak up river to Covington (directly across from where you started originally), how long will it take you? ans: ______sec

b. If you now paddle directly back across the river (back to Cincinnati), what angle (relative to the direct path across the river - the so-called crab angle) should you point the kayak to make it directly across? ans: ______ degrees

a. What is the velocity with which the block leaves the table? (Assume it leaves completely horizontally with no vertical component to the velocity)

b. What is the acceleration of the block (assume the +x direction is positive) while it is sliding along the table?

please show work.

thanks!

Explanation / Answer

1) lets say we row at theta with vertical.

So 2.4sin(theta)=0.9

so theta)=22 degrees(approx.)

an t=800/(2.4cos(theta))=359.6seconds

2)1.4=0.5*9.8*t^2

So t=sqrt(14)/7=0.5345 sec

So v=3.2/t= 6m/s(approx.)

a=-0.3*0.5*9.8=-1.47m/s^2(-ve means retardation)

6^2-u^2=2*a*1.4=-4.116

So u=6.33m/s

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