Finding the Direction and Distance of a Train's Motion Using Basic Kinematics

[miglo]

Homework Statement

A train moving with a constant velocity travels 135 m north in 12 s and an undetermined distance to the west. The speed of the train is 26 m/s.

(a) Find the direction of the train's motion relative to north. (in degrees west of north)

(b) How far west has the train traveled in this time?

Homework Equations

speed=distance/time or distance =speed*time

v_0cos(theta)=v_x
v_0sin(theta)=v_y

The Attempt at a Solution

i have no idea how to start, but I am guessing i have to find the distance traveled north first in order to find the angle theta.

 

[PeterO]

Homework Statement

A train moving with a constant velocity travels 135 m north in 12 s and an undetermined distance to the west. The speed of the train is 26 m/s.

(a) Find the direction of the train's motion relative to north. (in degrees west of north)

(b) How far west has the train traveled in this time?

Homework Equations

speed=distance/time or distance =speed*time

v_0cos(theta)=v_x
v_0sin(theta)=v_y

The Attempt at a Solution

i have no idea how to start, but I am guessing i have to find the distance traveled north first in order to find the angle theta.

You were given the distance North - You have to find the speed North.

 

[Rayquesto]

oh! ok so this problem is forcing you to calculate the x component of the velocity since they pretty much give you the y component to find theta.

Do you know how to do that?

 

[PeterO]

oh! ok so this problem is forcing you to calculate the x component of the velocity since they pretty much give you the y component to find theta.

Do you know how to do that?

Well it did travel 135m in 12 seconds.

 

[Rayquesto]

yes and that would be y. what's x?

 

[PeterO]

yes and that would be y. what's x?

Well it is logically negative, since the train went West, and the application of the Pythagorus theorem to the x and the y yields 26.

 

[Rayquesto]

thats how I see it too. The hidden part of the problem that will get you the answer lies in the pythagorean theorem.

 

[miglo]

ok i think I am finally understanding this
so since it says the train traveled 135 m in 12 s, then its velocity is 135m/12s right?
so i get 11.25m/s so that's the y component of velocity V_y
so since its overall speed is 26 m/s then by using the pythagorean theorem we can find V_x?
26=sqrt(V^2_x+V^2_y)=sqrt(V^2_x+(11.25)^2)
from this i got V_x=23.44m/s
so then tan(theta)=V_y/V_x=11.25/23.44
theta=arctan(11.25/23.44)=25.6 degrees
did i do this right?

edit: so i submitted my answer (my homework is online) and i got it wrong, i still have 3 more tries i believe
where did i go wrong?

 

[PeterO]

ok i think I am finally understanding this
so since it says the train traveled 135 m in 12 s, then its velocity is 135m/12s right?
so i get 11.25m/s so that's the y component of velocity V_y
so since its overall speed is 26 m/s then by using the pythagorean theorem we can find V_x?
26=sqrt(V^2_x+V^2_y)=sqrt(V^2_x+(11.25)^2)
from this i got V_x=23.44m/s
so then tan(theta)=V_y/V_x=11.25/23.44
theta=arctan(11.25/23.44)=25.6 degrees
did i do this right?

edit: so i submitted my answer (my homework is online) and i got it wrong, i still have 3 more tries i believe
where did i go wrong?

If you sketch a North component of 11.25 and a west component of 23.44, with resultant 26, you will see the triangle and perhaps recognise the 25.6 degree angle you quoted. But that is not the angle "west of North" as asked.

 

[Rayquesto]

Yes you are doing it right however the angle you found is relative to the west. How would you go about finding the angle relative to the north? (hint: you are very close and 90 degree angles determine frequencies in how one should go about calling the direction north, west, east, or south). I'm sorry if that sounded a bit wordy.