1. The acceleration is the derivative of the velocity.
Part (a)
The acceleration is the derivative of the velocity.
v(t) = 1 - tan-1(et)
a(t) = et e2t
a(2) = e e4
a(2) = or -.133

2. Since both v(2) and a(2) are negative, the speed is increasing.
Part (b)
v(2) = 1 - tan-1(e2)
v(2) =
a(2) = from Part (a)
Since both v(2) and a(2) are negative, the speed is increasing.
Remember from earlier in the year that speed INCREASES when v(t) and a(t) have the SAME sign, and speed DECREASES when the signs are OPPOSITE.

3. Part (c) v(t) = 1 - tan-1(et) 0 = 1 - tan-1(et) tan-1(et) = 1
The particle reaches its highest point when v = 0.
v(t) = 1 - tan-1(et)
0 = 1 - tan-1(et)
tan-1(et) = 1
tan 1 = et
et = tan 1
ln (et) = ln (tan 1)

4. Part (c)
Radians required here!!!
t (ln e) = ln (tan 1)
t = ln (tan 1)
t = .443 sec
ln (et) = ln (tan 1)

5. Part (c)
To justify that t=.443 is the highest point, we’ll find the velocity somewhere below and above t=.443 seconds.
v(.3) = 1 - tan-1(e.3)
= .067
t = .443 sec
v(.5) = 1 - tan-1(e.5) =
Since v(t) is positive for t<.443 and v(t) is negative for t>.443, the particle’s graph has an absolute maximum at t=.443

6. Part (d) v(t) = 1 - tan-1(et) Integrate both sides
y(2) = [1- tan-1(et)] dt 2
y(2) = or
The particle is moving AWAY from the origin since v(2) and y(2) are both negative.