Please Help... A robot travels at a speed of 1.35 m/sec at a direction of 45.0 degrees for 305 seconds. It then travels at a speed of 1.50 m/sec at a direction of 140.0 degrees for 852 seconds. What is the total displacement at the end of the second maneuver?

First, give the y component of the first displacement vector calculated in the problem.

Question

Please Help... A robot travels at a speed of 1.35 m/sec at a direction of 45.0 degrees for 305 seconds. It then travels at a speed of 1.50 m/sec at a direction of 140.0 degrees for 852 seconds. What is the total displacement at the end of the second maneuver? 

First, give the y component of the first displacement vector calculated in the problem.

anonymous · Answer

1308.08m
u can use the law of cosines
c^2=a^2+b^2-2ab cos (c) where cos(c)=cos(85degree)

anonymous · Answer

Is 1308.08 the y component or the total displacement?

anonymous · Answer

the total

anonymous · Answer

Firstly, I'm assuming that the degrees are calculated from a absolute degree frame. i.e. North is 000, East is 090. If not, please tell.
The first , y-component, can be calc'ed by this:
$S_y=u_y t + \frac{1}{2} a_y t^2$
Now, since $u_y$ will be $u sin 45^o$

anonymous · Answer

Then, a_y= 0, t=305, sub it all in to find the $S_y$ which is the y displacement.

anonymous · Answer

Do you need further help?