MA 3972-MA-Book April 11, 2018 15:14216 STEP 4. Review the Knowledge You Need to Score High
Solution:
(a) a(t)=v′(t) andv′(t) is the slope of the tangent line to the graph ofv.Att=1 andt=3,
the slope of the tangent is 0.
(b) For 2<t<4,v(t)>0. Thus, the particle is moving to the right during 2<t<4.
(c) Speed=|v(t)|att=1,v(t)=−4.
Thus, speed att=1is|− 4 |=4, which is the greatest speed for 0≤t≤4.TIP • Use only the four specified capabilities of your calculator to get your answer: plot-
ting graphs, finding zeros, calculating numerical derivatives, and evaluating definite
integrals. All other built-in capabilities can only be used tocheckyour solution.Vertical Motion
Example
From a 400-foot tower, a bowling ball is dropped. The position function of the bowling
balls(t)=− 16 t^2 +400,t≥0 is in seconds. Find:(a) the instantaneous velocity of the ball att=2 seconds.
(b) the average velocity for the first 3 seconds.
(c) when the ball will hit the ground.Solution:
(a) v(t)=s′(t)=− 32 t
v(2)=32(2)=−64 ft/sec
(b) Average velocity=
s(3)−s(0)
3 − 0=
(−16(3)^2 +400)−(0+400)
3
=−48 ft/sec.
(c) When the ball hits the ground,s(t)=0.
Thus, sets(t)= 0 ⇒− 16 t^2 + 400 =0; 16t^2 =400;t=±5.
Sincet≥0,t=5. The ball hits the ground att=5 sec.TIP • Remember that the volume of a sphere isv=^4
3
πr^3 and that the surface area iss= 4 πr^2.
Note thatv′=s.Horizontal Motion
Example
The position function of a particle moving in a straight line iss(t)=t^3 − 6 t^2 + 9 t−1,t≥0.
Describe the motion of the particle.Step 1: Findv(t) anda(t).
v(t)= 3 t^2 − 12 t+ 9
a(t)= 6 t− 12