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Woolyabyss
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This isn't really homework its an example, But I need help understanding it to do my homework
A destroyer is 500 km due west of a frigate. The destroyer is traveling at 10 km/h in a direction of 30 degrees north of east.The frigate is traveling at 5(2)^(1/2) in a NW direction.
(i) Find the velocity of the frigate relative to the destroyer.
(ii) Show that they are on a collision course.
(iii) When will they collide?
Vab = Va -Vb (V is a vector I don't know how to write vector notation on a computer)
d is the destroyer
f is the frigate
(solution of example as in the book)
(i)Vd = 10cos30i + 10sin30j
= 8.66i + 5j
Vf = -5(2)^(1/2)cos45i + 5(2)^(1/2)sin45j = -5i +5j
Vfd = (-5i +5j) - (8.66i +5j)
=-13.66i
(ii)
Position of frigate relative to the destroyer is at 500i km
we write this as Rfd = 500i km
the velocity of the frigate relative to the destroyer is
Vfr = -13.66i km/h (I think this is an error and it should Vfd)
since Vfd = -k(Rfd) where k is a positive constant , they must be on a collision course.
(iii)
the time of the collision is given by relative distance/ relative speed
500/13.66 = 36 hours and 36 minutes later.I understand (i) completely my main problem lies with (ii) where it says
"since Vfd = -k(Rfd) where k is a positive constant , they must be on a collision course."
The only way I can see they could make them both equal is if k were the inverse of time and if it is then why not just put k on the left and let it equal to time. I also don't get why the negative sign is needed.If anybody could help explain this one line to me it would be greatly appreciated.
Homework Statement
A destroyer is 500 km due west of a frigate. The destroyer is traveling at 10 km/h in a direction of 30 degrees north of east.The frigate is traveling at 5(2)^(1/2) in a NW direction.
(i) Find the velocity of the frigate relative to the destroyer.
(ii) Show that they are on a collision course.
(iii) When will they collide?
Homework Equations
Vab = Va -Vb (V is a vector I don't know how to write vector notation on a computer)
d is the destroyer
f is the frigate
The Attempt at a Solution
(solution of example as in the book)
(i)Vd = 10cos30i + 10sin30j
= 8.66i + 5j
Vf = -5(2)^(1/2)cos45i + 5(2)^(1/2)sin45j = -5i +5j
Vfd = (-5i +5j) - (8.66i +5j)
=-13.66i
(ii)
Position of frigate relative to the destroyer is at 500i km
we write this as Rfd = 500i km
the velocity of the frigate relative to the destroyer is
Vfr = -13.66i km/h (I think this is an error and it should Vfd)
since Vfd = -k(Rfd) where k is a positive constant , they must be on a collision course.
(iii)
the time of the collision is given by relative distance/ relative speed
500/13.66 = 36 hours and 36 minutes later.I understand (i) completely my main problem lies with (ii) where it says
"since Vfd = -k(Rfd) where k is a positive constant , they must be on a collision course."
The only way I can see they could make them both equal is if k were the inverse of time and if it is then why not just put k on the left and let it equal to time. I also don't get why the negative sign is needed.If anybody could help explain this one line to me it would be greatly appreciated.
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