## pathosdebater Group Title 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. one year ago one year ago

1. Otonashi Group Title

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

2. pathosdebater Group Title

Is 1308.08 the y component or the total displacement?

3. Otonashi Group Title

the total

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 $$u sin 45^o$$

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