Question

A boat radioed a distress call to a Coast Guard station. At the time of the call, a vector A from the station to the boat had a magnitude of 45.0 km and was directed 15.0° east of north. A vector from the station to the point where the boat was later found is B = 30.0 km, 15.0° north of east.

How far did the boat travel from the point where the distress call was made to the point where the boat was found? In other words, what is the magnitude of vector C?
A)65.3 km
B)39.7 km
C)26.5 km
D)54.0 km
E)42.5 km

Answer 1

d = 39.7 km

Step-by-step explanation:

initial position of the boat is 45 km away at an angle of 15 degree East of North

so we will have

`r_1 = 45 sin15 \hat i + 45 cos15 \hat j`

`r_1 = 11.64 \hat i + 43.46\hat j`

after some time the final position of the boat is found at 30 km at 15 Degree North of East

so we have

`r_2 = 30 cos15\hat i + 30 sin15 \hat j`

`r_2 = 28.98\hat i + 7.76 \hat j`

now the displacement of the boat is given as

`d = r_2 - r_1`

`d = (28.98\hat i + 7.76 \hat j) - (11.64 \hat i + 43.46\hat j)`

`d = 17.34 \hat i - 35.7 \hat j`

so the magnitude is given as

`d = √(17.34^2 + 35.7^2)`

`d = 39.7 km`