Relative Velocity

[legend217 0]

http://www.cchms.moe.edu.sg/kwek_wei_hong/...%20Velocity.pdf

Somebody help me with question 4 please, showing me the workings. I'll be very grateful.

[nightcrawler1089 0]

Well, for the first part, just set it up as vector addition, and create a triangle.

In other words, one leg is vertical, and has a value of 15. Another is horizontal (for convenience and to make it easier conceptually, set it at the top of the "15" line segment, and have it extend to the left).

You're at the bottom of the "15" line segment. Draw the hypotenuse and figure out the angle and length of the hypotenuse.

Soulver (a Mac app) tells me the actual velocity is 66.708 km/h and it's headed northwest at 143.84 degrees.

Well, I'm pretty sure that's how the first part is done. However, I guarantee nothing .

-NC

P.S: The second part is somewhat similar, though more "confusing" because of the added time element. Just consider that it's one hour later, so the liner would have moved 25 km at 45º, and the tanker would have moved 15 km north. Set it up with vectors again.

[gnomexp 0]

Oh god.... geometry physics.

/goes to the corner and cries.

[legend217 0]

i can't get that 143.84 degrees. What i got was a bearing of approx 283 degrees.

And for the second part i got a strange quadilateral instead of a triangle. The dist apart i calculated is 29.8km but i'm not sure whether its correct.

Help pls. TIA

[zondajag 0]

wow i love this stuff...im going to study maths and physics at university next year.......but this question is rather trickey...i need a calculator...the second question seems to be a trick one to me as the swimmer can only swim at 5m/s in still water....in order for him to swim at this speed in the forward direction he must also be moving horizontally at the rate of the flow of current....maybe he needs to swim at a velocity equal and opposite in the direction of flow

[nightcrawler1089 0]

Yeah, my bad on the degree bearing--I meant to do arctan but accidentally took tan. This produces an angle measure of about 77 degrees, which you'll have to interpret yourself.

-NC

[thehundredthone 0]

for the first part I seem to be getting 66.708 kph at 12.99deg NW (angle between tanker and speedboat = 102.99deg)

I used relative velocity for this. given that the angle between the resultant and 1 velocity is pi/2 rad,

v1+v2.cos(theta) = 0

hence v2.cos(theta)=-v1

substituting this value in the eqn R^2 = v1^2 + v2^2 + 2.v1.v2cos(theta)

v2 comes to 66.708

also given that v1 + v2.cos(theta) = 0

cos(theta) = -v1/v2 => theta = 102.99deg

now if you draw a diagram with v1 as a vertical line, R (65kph) as a horizontal line (or vectors ) and v2 at an arbitrary angle, seeing as theta, the angle between v1 and v2 is 102.99, and the angle beta between v1 and R is 90deg or pi/2 rad; the angle between R and v2 is 12.99deg or 12.99deg N of W

for the second part

the tanker moves 15km north in 1 hr, however at the same time the liner moves 25/sqrt2 km north as well (the vertical component of its velocity which is 25sin(pi/4) ), so the vertical distance is 15-(25/sqrt2) = -2.68km, ie the liner is north of the tanker by 2.68 km

now the horizontal distance initially was 12km, and the liner moves 25/sqrt2 km east therefore the horizontal distance becomes 12+(25/sqrt2) km (horizontal component = 25cos(pi/4) = 29.68 km

taking these as the arms of a right angled triangle, the distance between them at 0900 hrs is 29.80 km