I am having trouble workings vector problems. Here isone that has me stuck... A
ID: 1763639 • Letter: I
Question
I am having trouble workings vector problems. Here isone that has me stuck... A car is traveling directlywest for 16km west at 90km/h, then it travels directlysouth for 8km at 80km/h, then it travels directly SE for 34km at100km/h. Assuming it does not stop or slow down when changingdirection, a) What is the change in velocity for thos tripaqnd b) what is its average acceleration? I am having trouble workings vector problems. Here isone that has me stuck... A car is traveling directlywest for 16km west at 90km/h, then it travels directlysouth for 8km at 80km/h, then it travels directly SE for 34km at100km/h. Assuming it does not stop or slow down when changingdirection, a) What is the change in velocity for thos tripaqnd b) what is its average acceleration?Explanation / Answer
The initial velocity of the car is 90km/h toward west Therefore vix = -90km/h and viy =0km/h Final velocity is 100km/h at South East. Therefore vfx= 100km/h cos45o =70.7km/h and vfy = -100km/h sin45o =-70.7km/h Then change in x-component of velocity is 70.7km/h- (-90km/h)= 160.7km/h and change in y - component of velocity is -70.7km/h therefore the change in velocityis [(160.7km/h)2 + (-70.7km/h)2]= 175.56km/h Direction is tan-1(-70.7/160.7) =23.7oSouth of East
Total time of travel is (60km/90km/h) + (8km/80km/h) +(34km/100km/h) = 1.106h Therefore the acceleration a = change in velocity /time =(175.56km/h)/1.106h =0.0122m/s2 at 23.7o South of East
Hope this helps you
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.