A dog running in an open field has components of velocity
vx = 2.6 m/s and vy = -1.8 m/s at t1 = 10.0s. For the time interval from t1 = 10.0 s to t2 = 20.0 s, the average acceleration of the dog has magnitude 0.45 m/s^2 and direction 31.0 degrees measured from the +x-axis toward the +y-axis. At t2 = 20.0s,
a) what are the x- and y-components of the dog's velocity?
b) what are the magnitude and direction of the dog's velocity?
a) This is a simple kinematics problem in two dimensions, but first we need to find the x- and y-components of the acceleration of the dog from 10 to 20s.
Since the angle is 31 degrees from the positive X toward the positive Y-axis, (quadrant 1), the ax and ay components are as simple as
ax = 0.45 * cos(31) = .386
ay = 0.45 * sin(31) = .232
Then, we can apply this directly to the kinematics equation for velocity,
v = v0 + a*t
for each of the components.
vx = 2.6 + .386*(10) = 6.46 m/s
vy = -1.8 + .232*(10) = 0.518 m/s
b) To find the magnitude and direction, we just convert the individual x and y-components into vector form.
The magnitude is found by the distance formula,
v = (vx^2 + vy^2)^(1/2)
v = (6.46^2 + 0.518^2)^(1/2) = 6.48 m/s
and the direction can be found using arctan(vy/vx) as we have both the x- and y- components.
arctan(0.518/6.46) = 4.57 degrees.
