easy peasy problems that i cant seem to get T_T (1 Viewer)

[jeniii]

hey guys i just started in a new skool this year for year 11 and im already struggling T_T i dont get all this relative velocity stuff. can someone please explain to me step by step (i.e. why its N bla bla W degrees) pretty please

1. by drawing a vector diagram, find the resultant displacement of a car, which drives 10 km N, then 15 km E then 7 km SW then 4 km N.

2. A car travels north at 60 kph for 2 hours. it then travels west for 5 hours at 80 kph. calculate average velocity. (i got like 120 + 400 + 58???)

3. A cyclist riding due west at 30 ms ^ -1 felt a breeze in his face at 32 ms^-1 from a direction W 10 degrees N. What was the velocity of the breeze relative to the ground?

omg.. i seriously dont even get wat question 3 is saying T_T can someone explain to me any of those questions? i actually have a lot more.. but mayb if i get how to do those i can do the rest

 

[Riviet]

3. A cyclist riding due west at 30 ms ^ -1 felt a breeze in his face at 32 ms^-1 from a direction W 10 degrees N. What was the velocity of the breeze relative to the ground?

omg.. i seriously dont even get wat question 3 is saying T_T can someone explain to me any of those questions? i actually have a lot more.. but mayb if i get how to do those i can do the rest

Question 1 is not easy, still trying to figure out how to find the displacement. For question 3, W 10^o N means the angle made from the west direction to the north direction which in this question is 10 degrees. The velocity of the breeze relative to the ground means how fast the breeze is in comparison to the ground (which is not moving or stationary). You need to use vector diagrams and subtract the velocity of the cyclist since he/she is riding against the breeze.

 

[speedie]

wow what school r u in now, its pretty hard stuff for yr 11

 

[airie]

hey guys i just started in a new skool this year for year 11 and im already struggling T_T i dont get all this relative velocity stuff. can someone please explain to me step by step (i.e. why its N bla bla W degrees) pretty please

1. by drawing a vector diagram, find the resultant displacement of a car, which drives 10 km N, then 15 km E then 7 km SW then 4 km N.

2. A car travels north at 60 kph for 2 hours. it then travels west for 5 hours at 80 kph. calculate average velocity. (i got like 120 + 400 + 58???)

3. A cyclist riding due west at 30 ms ^ -1 felt a breeze in his face at 32 ms^-1 from a direction W 10 degrees N. What was the velocity of the breeze relative to the ground?

omg.. i seriously dont even get wat question 3 is saying T_T can someone explain to me any of those questions? i actually have a lot more.. but mayb if i get how to do those i can do the rest

I drew up a diagram for q1 (attached) and used cosine rule and got something weird...hmmm.

For q2, I got sqrt(120^2 + 400^2) / (2+5) = (40*sqrt109) / 7 kph.

Hope that helps...and hope it's correct

 

[webby234]

2. displacement north = 120km
displacement west = 400km

total displacement = sqrt(400^2+120^2)
= sqrt174400
= 40rt109
velocity = 40rt109/7 km/h (can't be bothered to make it m/s)

tan angle = 400/120

73 degrees
360-73 = 287

so velocity is 40rt109/7km/h @ 287 degrees

 

[Mountain.Dew]

for q1 have a look at this diagram...

just by using pythagoras:

first find x, then find y, then find ur displacement, which is the distance from the lower left hand corner vertices of the triangle, to the top middle point which seperates x and y.

hope it helps, MD.

 

[vizman]

for q1 have a look at this diagram...

just by using pythagoras:

first find x, then find y, then find ur displacement, which is the distance from the lower left hand corner vertices of the triangle, to the top middle point which seperates x and y.

hope it helps, MD.

how can u assume the 4km connects with the other line

 

[Riviet]

Also MD, we can't assume that the third path meets up with the start again, you can see that it's not an isoscles triangle, since the two sides need to be equal for that to happen. If you were to form a triangle though, you would have to extend the starting point back by 5 km.

I drew it to scale, and the 4 side length doesn't meet up with the 15 side length, it's actually short by about 1 cm. I measured the displacement to be about 13.5 km.

 

[Riviet]

Here's my diagram for question 1. It's alot easier than what it looked beforehand. =D

 

[minushuman]

If you couldn't work out that in your head then you've obviously failed at life. Plz die.

 

[airie]

2. displacement north = 120km
displacement west = 400km

total displacement = sqrt(400^2+120^2)
= sqrt174400
= 40rt109
velocity = 40rt109/7 km/h (can't be bothered to make it m/s)

tan angle = 400/120

73 degrees
360-73 = 287

so velocity is 40rt109/7km/h @ 287 degrees

Oh yes, forgot the direction for the vector. My bad.

for q1 have a look at this diagram...

just by using pythagoras:

first find x, then find y, then find ur displacement, which is the distance from the lower left hand corner vertices of the triangle, to the top middle point which seperates x and y.

hope it helps, MD.

By using similar triangles we can actually find that the 4km line ends at a point north of the 15km line (excuse the diction ), as the distance between the point that marks 7km SW and the 15km line is 14/13 * sqrt13, found by using ratios between corresponding sides of similar triangles. I'm sure that cosine rule would find the answer, except...heh, I have no idea how to simplify a square root inside a square root sign