Ship A is located 4 km north and 3 km east of ship B. Ship A has a velocity of `20 kmh^-1` towards the south and ship B is moving at `40 kmh^-1` in a direction `37^@` north of east. X and Y-axes are along east and north directions, respectively

Options

A. Velocity of A relative to B is `-32hati - 44hatj`
B. Position of A relation to B as a function of time is given by
`vecr_(AB) = (3-32t)hati + (4-44)hatj`
where t = 0 when the ships are in position described above.
C. Velocity of B relative to A is ` - 32hati - 44hatj`.
D. At some moment A will be west of B.

Answer

A, B

Explanation

a,b. `vecv_(AW) = -20hatj`
`vecv_(BW) = -32hati + 24hatj`
`vecv_(AB) = vecv_(AW) - vecv_(BW) `
` = -32hati - 44 hatj`
`(dvecr_(AB))/(dt) = vecv_(AB)`
`= - 32hati - 44hatj`
`int_(3hati+4hatj)^(vecr_(AB)) d(vecr_(AB)) = - int_(a)^(t) 32 dt hati - int_(0)^(t) 44 dthatj`
`vecr_(AB) = (3-32t)hati + (4-44t)hatj`
At `t = 1//11h,` A will be east of B. So at no time, A will
component is `3 - 32t = 3 - 32/11 = 1/11 km`.
It means at `t = 1/11h`, A will be east of B. So at no time, A will
be west to B.