a. add the two velocities as vectors.
find the angle of travel down stream
b. distance downstream: you have the angle, and the distance across
c. time=250m/3.20m/s
d. distance downstream=timeinwater/1.20
A boat sets off perpendicular to the bank, to cross a river width of 250m. The speed of the boat in still water is 3.20ms-1, however there is a current downstream at 1.20ms-1, which puss the boat sideways.
a) Find the resultant velocity of the boat.
b) Find the total distance the boat travels in crossing the river.
c) How long does the crossing take?
d) How far downstream from the starting point does the boat travel?
4 answers
thanks. the answers to b. is 167m and the answer to d. is 93.7 m. but I'm not sure how to get that. would you please care to explain how to get that?
b. angle: arctan 1.2/3.2
then distance downstreem=250/cosangle
I get for the angle...0.359radians
then for distance = 250/cos.259=259m
Your answer is not possible.
Now another way: time to get across=250/3.2=78.1sec
distance downstream=78.1*1.2=93.7m
now distance total=sqrt(93.7^2+250^2) which is greater than your answer also.
your answer to b is wrong.
then distance downstreem=250/cosangle
I get for the angle...0.359radians
then for distance = 250/cos.259=259m
Your answer is not possible.
Now another way: time to get across=250/3.2=78.1sec
distance downstream=78.1*1.2=93.7m
now distance total=sqrt(93.7^2+250^2) which is greater than your answer also.
your answer to b is wrong.
a. Vb + Vc = 3.2i + 1.2 = 3.42m/s[69.4o]
= 3.42m/s[20.6o] E. of N.
b. d = 250/Cos20.6 = 267 m.
c. d = V*t = 267 m
3.42*t = 267
t = 78.1 s.
d. d = 250*Tan20.6 = 94 m
= 3.42m/s[20.6o] E. of N.
b. d = 250/Cos20.6 = 267 m.
c. d = V*t = 267 m
3.42*t = 267
t = 78.1 s.
d. d = 250*Tan20.6 = 94 m