A wild horse starts from rest and runs in a straight line 29° north of west. After 39 s of running in this direction, the horse has a speed of 10 m/s.
(a) What is the magnitude of the horse's average acceleration?
(b) Assuming that north and east are the positive directions, find the component of the horse's acceleration that points along the north-south line.
(c) Find the component of the horse's acceleration that points along the east-west line.

Question

anonymous · Answer

I know that the answer to a) is .26 m/s^2

anonymous · Answer

acc = change in velocity per unit time.

anonymous · Answer

That doesn't help at all. I have no clue how to solve it

anonymous · Answer

acc=change in velocity/time taken=(10-0)/39=?

anonymous · Answer

acceleration = change in velocity / total time taken

= 10 / 39 m/s^2 = 0.2564 m/s^2

|dw:1316702226674:dw|
if u resolve the vectors along the direction of x and y axis then u will get
acceleration in y direction to be 0.2564 sin 29=

0.2564 * 0.484 = 0.1240 m / s^2


similarly acceleration along y direction = 0.2564 * cos 29

= =0.2564 * 0.874 = 0.2240 m/s^2