Velocity/direction and vectors

[nick901304]

Homework Statement

An off-roader explores the open desert in her Hummer. First she drives 25 degrees west of north with a speed of 6.5 m/h for 15 minutes, then due east with a speed of 11m/h for 7.5 minutes. She completes the final leg of her trip in 22 minutes.

Part A
What is the direction and speed of travel on the final leg? (Assume her speed is constant on each leg, and that she returns to her starting point at the end of the final leg.)
Express your answer using two significant figures.
ANSWER: = ?
degrees south of west



Part B
Express your answer using two significant figures.
Express your answer using two significant figures.
ANSWER: = ? m/h




I don't even know where to start. I tried drawing it out then using trig to get the angles (my answer didnt work). I also used pythag theorum to get the other side of the triangle she makes with her vehicle, but that answer didnt work either >_< would love some help, or even just a explanation to where to start the problem

 

[Nabeshin]

None of your speeds have any units on them... but whatever.

Perhaps it would help if you were to break this down into three separate sets of motion:
1st while she is driving 25º W of N (find out the x and y displacements)
2nd, while she is driving E (the same)
And finally, the last leg.

You should know now how far from the starting point she is and assuming she drives in a straight line draw a picture to that effect. This should help :)

 

[nlightner]

.

I would approach this problem by first identifying the known and unknown variables. The known variables in this problem are the initial direction (25 degrees west of north) and speeds (6.5 m/h and 11 m/h) for the first two legs of the trip, as well as the time durations of those legs (15 minutes and 7.5 minutes). The unknown variables are the direction and speed of the final leg.

To solve for the direction and speed of the final leg, we can use vector addition. This involves breaking down each leg into its x and y components, and then adding them together to get the final vector.

For the first leg, the x component is -6.5 cos(25) m/h and the y component is 6.5 sin(25) m/h. Similarly, for the second leg, the x component is 11 cos(0) m/h and the y component is 11 sin(0) m/h.

To find the final direction and speed, we can add the x and y components of the first two legs together, and then use the inverse tangent function to find the angle and the Pythagorean theorem to find the speed.

The x component of the final leg would be the sum of the x components of the first two legs, which is -6.5 cos(25) + 11 cos(0) = 3.94 m/h. The y component of the final leg would be the sum of the y components of the first two legs, which is 6.5 sin(25) + 11 sin(0) = 4.38 m/h.

Using the inverse tangent function, we can find the angle of the final leg, which is tan^-1(4.38/3.94) = 50.3 degrees. This means that the final leg is 50.3 degrees south of west.

To find the speed of the final leg, we can use the Pythagorean theorem, which states that the magnitude of a vector is equal to the square root of the sum of the squares of its components. In this case, the magnitude of the final leg vector is:

√(3.94^2 + 4.38^2) = 5.7 m/h.

Therefore, the final leg of the trip has a speed of 5.7 m/h and is 50.3 degrees south of west.

In