AP Calc Notes: DI – 10 Position, Velocity and Acceleration

AP Calc Notes: DI – 10 Position, Velocity and Acceleration
Calculus: The Musical –Physics Extravaganza lyrics
Recall:
Position
m
s (t )
Velocity
m/s
s ' (t )
v (t )
Acceleration
m/s2
s "(t )
v '(t )
a (t )
Given v(t)
Formula
b
Change in position on [a, b]
∫ v ( t ) dt = s ( b ) − s ( a )
a
b
s ( b ) = s ( a ) + ∫ v ( t ) dt
a
Position at t = b
b
∫ v ( t ) dt
a
Total distance traveled on [a, b]
Average acceleration at t = a
v (b ) − v ( a )
b−a
Acceleration at t = a
v '(a)
Average velocity on [a, b]
b
s (b ) − s ( a )
1
v ( t ) dt =
∫
b−a a
b−a
Speed
v (t )
Recall:
Speed is increasing if velocity and acceleration have the same sign
Speed is decreasing if velocity and acceleration have different signs
Units
With Calculator
A bug ambles back and forth along the x-axis with a velocity in cm/s given by v ( t ) = 0.4t 3 − 2t 2 + t + 2 for all
time t ≥ 0. At time t = 0, the bug is at x = 1. Find the following:
a. v(2) = 0.4(2)3 - 2(2)2 + 2 + 2 = -0.8 cm/s
b. a(2) = v'(2) = -2.200 cm/s2 (Use nDeriv.)
c. Is the bug speeding up or slowing down at time t = 2?
d. x(2) = x(0) +
e. x(5) = x(0) +
∫
2
∫
5
0
0
v(t)dt = 1 + 2.267 = 3.267 cm
v(t)dt = 1 + 1.667 = 2.667 cm (Use fnInt.)
f. Average velocity from t = 2 to t = 5:
One way: v =
OR: v =
5
1
v(t)dt = -0.200 m/s
∫
5-2 2
y
Δx
3.267 - 2.667
=
= -0.2 m/s
Δt
3
g. Total distance traveled from t = 2 to t = 5:
dis tan ce =
∫
5
2
x
| v(t) | dt = 6.103 cm
Bug travels
backward.
Bug travels
forward.
b
Change in velocity from t = a to t = b
∫ a ( t ) dt = v ( b ) − v ( a )
a
b
Change in position from t = a to t = b
∫ v ( t ) dt = s ( b ) − s ( a )
a
Position at time t = b, given v ( t ) and position at time t = a
b
∫ v ( t ) dt = s ( b ) − s ( a )
a
Average rate of change of velocity from t = a to t = b?
Know velocity
v (b ) − v ( a )
avg acceleration =
b−a
b
s ( a ) + ∫ v ( t ) dt = s ( b )
a
Know acceleration but not velocity
b
avg acceleration =
1
b−a
∫ a ( t ) dt = ( v ( b ) − v ( a ) )
1
b−a
a
Ex: Velocity of a particle is given by v ( t ) = 2t + 1 .
a. Find the change in position for the particle on [0, 3].
b. Find the average velocity of the particle on [1, 2].
c. Find the average acceleration of the particle on [1, 2].
d. Find the acceleration of the particle at t = 3.
e. Find the position of the particle at 5 seconds if the particle was 10 meters from the origin at 2 seconds.