2.2b Rates of Change

Position, velocity, acceleration

The position function (usually called s) relates position to time.You might be familiar with

s(t) = -16t2 + v0t + s0

from physics or even algebra 2

This function is used when distance is measured in feet and v0 is a fancy way of writing “velocity at time zero, or initial velocity” and s0 is a fancy way of writing “position at time zero, or initial position”.

Let’s look at s(t) = -16t2 - 30t + 150Velocity to start is 30 ft/sec (someone pushed you!), you are falling from a cliff 150 feet high, and you want to know your position at different times. s(0) = ? s(1) = ? s(2) = ?How long till you splat on the ground?

s(0) = 150 ft

s(1) = -16 - 30 + 150 = 104 ft

s(2) = -64 - 60 + 150 =36 ft

Hitting the ground means position from the ground is 0, so solve

20 16 30 150t t t=2.26 or -4.14 seconds

Change in DistanceAverage velocity=

Change in Time

s

t

DistanceRate=

Time

Ex. 9 p. 113 If a billiard ball is dropped from a height of 100 feet its position is given by s(t) = -16t2 +100.Find the average velocity over the following intervals.a. [1, 2]

b. [1, 1.5]

c. [1, 1.1]

a. Δs = s(2) – s(1) = 36 – 84 = -48 avg vel = -48/1 = -48 ft/sec

b. Δs = s(1.5) – s(1) = 64 – 84 = -20 avg vel = -20/.5= -40 ft/sec

c. Δs = s(1.1) – s(1) = 80.64 – 84 = -3.36 avg vel = -3.36/.1= -33.6 ft/sec

Note that the average velocities are negative, indicating downward movement

If you want to find instantaneous velocity at a moment of time, you would want to calculate the average velocity as the change in time got smaller and smaller. In other words, find the rate of change (slope) of the position function at time t. Sounds like a derivative idea!

0

( ) ( )( ) lim '( )

t

s t t s tv t s t

t

The velocity function is the derivative of the position function!

 

Acceleration is the derivative of velocity with respect to time, so 2

2( ) ( )

dv d va t v t

dt dt

20 0

1( )

2s t gt v t s

In general, the position function for a falling object is given as

where s0 is initial position, v0 is initial velocity, and g is the acceleration due to gravity. On earth that is -32 feet per second per second or -9.8 meters per second per second.

Ex 10 p. 114 Using the derivative to find velocity

At time t=0, a diver jumps from a platform diving board that is 32 feet above the water. The position of the diver is given by

a. When does the diver hit the water?b. What is the diver’s velocity on impact?

2( ) 16 16 32s t t t

a. Solve for s = 0. 20 16 16 32t t Hint: solve(expression, variable, guess, {lower, upper}) or solver, or polysmlt app! Or quadratic formula!t = -1 or 2

b. Find s’(2): '( ) 32 16s t t '(2) 32 2 16 48 / secs ft

35

30

25

20

15

10

5

s x = -16x2+16x+32

(0.5, 36)

Notice the diver moves upward for the first half second, because the velocity is positive for the first half second. When the velocity is zero, the diver has reached the maximum height of the dive.

Particle Motion

A particle moves along a line so that its position at any time t 0 is given by the function s(t) = t2 – 4t + 3, where s is in meters, t is in seconds.a) Find the displacement of the particle during the first 2 secs.

Sol: The displacement is given by s(2) – s(0) = (-1) – 3 = -4

b) Find the average velocity of the particle during the first 4 secs

Sol: Avg velocity is (4) (0) 3 30 m/sec

4 0 4

s s

c) Find the instantaneous velocity of the particle when t = 4

Sol: Instantaneous velocity if just the velocity v(t) where v(t) = ds/dt = 2t – 4. So v(4) = 4 m/sec

d) Find the acceleration of the particle when t = 4

Sol: Acceleration a(t) at any time t is a(t) = dv/dt = 2 m/sec2

 

For 0 t 2, v(t) < 0, so the particle is moving to the left. (Notice s(t) is decreasing). The particle starts (t = 0) at s = 3 and moves left, arriving at the origin t = 1when s = 0. It continues moving left until it reaches the point s = -1 at t = 2. At t = 2, v = 0, so the particle is at rest.For t >2, v(t) > 0, so particle is going to right. (Notice s(t) is increasing). In this interval, the particle starts at s = -1, moving to the right through the origin and continuing to the right for the rest of the time.The particle changes direction at t = 2 when v = 0.

f) Use parametric graphing to view the motion of the particle on the horizontal line y = 2Sol: Change mode to PAR (parametric)

EnterAnd graph in window [-5, 5] by [-2, 4] with Tmin = 0, Tmax = 10 (could go to infinity) and Xscl = Yscl= 1. Use trace to follow the path of the particle.

If you change your Tstep = 0,5, you see what is happening every half second.

21TX 4 3,T T 1T 2,Y

2.2b Assignment: p. 115 #75-76, 83-90, 93-96, 103-104 and Particle Motion Prob 1.