A wild horse starts from rest and runs in a straight line 29° north of west. After 39 s of running in this dir?
Please help. tnkz <3
Update: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.
What is the magnitude of the horse's average acceleration?
Assuming that north and east are the positive directions, find the component of the horse's acceleration that points along the north-south line
Find the component of the horse's acceleration that points along the east-west line.
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Verified answer
For the case when the starting speed is zero use the following equation.
a = v/t
where
a = average acceleration
t = time
v = velocity
so
a = (10 m/s) / 39 s
a = 0.26 m/s^2
Vy = component of acceleration along N-S line
Vy = 0.26 m/s^2 sin 39
Vy = 0.16 m/s^2
Vx = component of acceleration along E-W line
Vx = - 0.26 m/s^2 cos 39
Vx = -0.20 m/s^2
You raise some good points here.