A dog running in an open field has components of velocity

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.

For a system in simple harmonic motion, which of the following is the time required to complete a cycle of motion?

a. frequency b. amplitude c. period d. revolution

The answer is c. the period.

What does the length of a force vector represent?

The length of a force vector represents the magnitude of that force.

A newton is equivalent to which of the following quantities?

a.	kg	c.	kg·m/s^2
b.	kg·m/s	d.	kg·(m/s)^2

The answer is c. kg*m/s^2.  Newtons are the SI unit of force (units of mass*distance/time^2).