MA 3972-MA-Book April 11, 2018 15:14214 STEP 4. Review the Knowledge You Need to Score High
Sincef′(x)=cosxandf′(
π
6)
=cos(
π
6)
=√
3
2, you can use linear approximations:f(
π
6+
π
180)
≈ f(
π
6)
+f′(
π
6)(
π
180)≈sin
π
6+
[
cos(
π
6)] (
π
180)≈
1
2
+
√
3
2(
π
180)
= 0 .515.10.3 Motion Along a Line
Main Concepts:Instantaneous Velocity and Acceleration, Vertical Motion,
Horizontal MotionInstantaneous Velocity and Acceleration
Position Function: s(t)
Instantaneous Velocity: v(t)=s′(t)=
ds
dt
If particle is moving to the right→, thenv(t)>0.
If particle is moving to the left ←, thenv(t)<0.
Acceleration: a(t)=v′(t)=
dv
dt
ora(t)=s′′(t)=
d^2 s
dt^2
Instantaneous Speed: |v(t)|Example 1
The position function of a particle moving on a straight line iss(t)= 2 t^3 − 10 t^2 +5. Find
(a) the position, (b) instantaneous velocity, (c) acceleration, and (d) speed of the particle at
t=1.
Solution:
(a) s(1)=2(1)^3 −10(1)^2 + 5 =− 3
(b) v(t)=s′(t)= 6 t^2 − 20 t
v(1)=6(1)^2 −20(1)=− 14
(c) a(t)=v′(t)= 12 t− 20
a(1)=12(1)− 20 =− 8
(d) Speed=|v(t)|=|v(1)|= 14Example 2
The velocity function of a moving particle isv(t)=
t^3
3− 4 t^2 + 16 t−64 for 0≤t≤7.