With acceleration

and initial velocity

the velocity at time <em>t</em> (b) is given by




We can get the position at time <em>t</em> (a) by integrating the velocity:

The particle starts at the origin, so
.



Get the coordinates at <em>t</em> = 8.00 s by evaluating
at this time:


so the particle is located at (<em>x</em>, <em>y</em>) = (64.0, 64.0).
Get the speed at <em>t</em> = 8.00 s by evaluating
at the same time:


This is the <em>velocity</em> at <em>t</em> = 8.00 s. Get the <em>speed</em> by computing the magnitude of this vector:

Work = (force) x (distance) =
(200 N) x (3.5 m) = <em>700 joules</em>
The woman is correct because it is the pace he is moving not walking.
This problem here is an example of inelastic collision where kinetic energy is not conserved but momentum is. We calculate as follows:
m1v1 + m2v2 = (m1 + m2)v3
v3 = m1v1 + m2v2 / m1 + m2
v3 = (30.2)(1000) + (5000)(0) / (30.2 + 5000)
v3 = 6.00 m/s