1. VELOCITY, ACCELERATION & RATES OF CHANGE

2. Velocity and acceleration—rates of change
Velocity is the “average” speed
Acceleration is the speed at an instant
If it is 60 miles to Jacksonville and you drive it in an hour, your “average” speed was 60 mph
If you looked at your speedometer at any time during the trip, your speed at that moment was your instantaneous speed.
Acceleration is the derivative of velocity, or how fast your speed is changing

3. 𝑓 𝑡 = 49ℎ+ 4.9ℎ 2 ℎ (now take the limit as h->0)
A BALL IS DROPPED FROM A TOWER 450 METERS HIGH. THE POSITION FUNCTION IS 𝒔=𝒇 𝒕 =𝟒.𝟗 𝒕 𝟐 . (a) find the speed after 5 sec (find the slope at 5 sec). (b) what is the speed when the ball hits the ground (find the derivative and then find the speed)
𝑓 𝑡 = 4.9∗ 5+ℎ 2 −4.9∗ 5 2 ℎ
𝑓 𝑡 = 4.9∗ 25+10ℎ+ ℎ 2 −4.9∗25 ℎ
𝑓 𝑡 = 49ℎ+ 4.9ℎ 2 ℎ (now take the limit as h->0)
𝑓 𝑡 = lim ℎ→0 49ℎ+4.9 ℎ 2 ℎ =49 ( 𝑚 𝑠 )

4. Part b—find the derivative first, then find the speed
𝑠=𝑓 𝑡 = 4.9∗ 𝑥+ℎ 2 −4.9 𝑥 2 ℎ
𝑠=𝑓 𝑡 = 4.9 𝑥 𝑥ℎ+4.9 ℎ 2 −4.9 𝑥 2 ℎ
𝑠=𝑓 𝑡 = 9.8𝑥ℎ+4.9 ℎ 2 ℎ (take the limit as h->0)
lim ℎ→ 𝑥ℎ+4.9 ℎ 2 ℎ =9.8𝑥
𝑓 ′ 𝑡 =9.8𝑥 which is a generic equation to find the speed for any time in our interval

5. now find the time the object falls
4.9 𝑡 2 =450𝑚
𝑡= = 9.6𝑚 𝑠
Plug in this value into the ‘generic’ equation 𝑓 ′ 𝑡 =9.8𝑥
𝑓 ′ 𝑡 =9.8∗9.6=94 𝑚 𝑠

6. ANOTHER ONE
A ball is thrown into the air with a velocity of 40 ft/sec and it’s position ( height) is given by the equation 𝑦=40𝑡−16 𝑡 2 . Find the speed at 2 seconds (do the derivative first)
𝑦= 40 𝑥+ℎ −16 𝑥+ℎ 2 − 40𝑥−16 𝑥 2 ℎ
𝑦= 40𝑥+40ℎ−16 𝑥 2 −32𝑥ℎ−16 ℎ 2 −40𝑥+16 𝑥 2 ℎ
𝑦= 40ℎ−32𝑥ℎ−16 ℎ 2 ℎ (take the limit as h->0)
y ′ =lim ℎ→0 40ℎ−32𝑥ℎ−16 ℎ 2 ℎ =40−32𝑥

7. 𝑦 ′ =40−32𝑥
Substitute in 2 seconds for x
𝑦 ′ =40−64
𝑦 ′ =−24 𝑓𝑡/𝑠𝑒𝑐

8. One more
The displacement for a particle moving in a straight line is given by the equation s= 1 𝑡 Find the velocity of the particle at 1 and 3 seconds.

9. s= 1 𝑡 2 .
𝑠= 1 𝑥+ℎ 2 − 1 𝑥 2 ℎ
𝑠= 1 𝑥 2 +2𝑥ℎ+ ℎ 2 − 1 𝑥 2 ℎ
𝑠= 𝑥 2 − 𝑥 2 −2𝑥ℎ− ℎ 2 ℎ∗ 𝑥 2 ∗( 𝑥 2 +2𝑥ℎ+ ℎ 2 )
lim ℎ→0 𝑠 𝑡 = −2𝑥+ℎ 𝑥 2 ∗ 𝑥 2 +2𝑥ℎ+ ℎ (now take lim h->0)
𝑠=− 2𝑥 𝑥 4
𝑠=− 2 𝑥 3

10. 𝑠=− 2 𝑥 3
At 1 second s = -2
At 3 seconds 𝑠=− 2 27