Question #4257

you are able to swim at 1.4 m/s. one day you swim directly across a river that has a uniform water flow of 6.1 m/s. what is your resultant swimming velocity - both magnitude and direction (given as an angle with respect to the shore)?

Expert's answer

Let V be the vector of my velocity, and U be the vector of water flow

velocity.

By assumption V and U are orthogonal, |V|=1.4, and

|U|=6.1

If we choose plane coordinates (x,y) in which I have coordinate

(0,0), V=(0, 1.4) directed up, and U=(6.1, 0) directed

right, then the

resulting vector is

V+U = (6.1, 1.4)

It magnitude is equal to

|V+U| = sqrt( 6.1 * 6.1& +& 1.4 * 1.4 ) = sqrt(39.17) ~ 6.2586

m/s

Its angle with respect to the shore is

phi = arrcos (6.1 /

6.2586) = arccos(0.97466) ~ 12.926 degrees

velocity.

By assumption V and U are orthogonal, |V|=1.4, and

|U|=6.1

If we choose plane coordinates (x,y) in which I have coordinate

(0,0), V=(0, 1.4) directed up, and U=(6.1, 0) directed

right, then the

resulting vector is

V+U = (6.1, 1.4)

It magnitude is equal to

|V+U| = sqrt( 6.1 * 6.1& +& 1.4 * 1.4 ) = sqrt(39.17) ~ 6.2586

m/s

Its angle with respect to the shore is

phi = arrcos (6.1 /

6.2586) = arccos(0.97466) ~ 12.926 degrees

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