Lesson VelocityPVA, derivatives, anti-derivatives, initial value problems, object moving left/right and at rest

Position, Velocity, and AccelerationPVA problemsPosition:

Velocity:

Acceleration:

Speed =

¿ 𝑠 ’ (𝑡 )

¿𝑣 ’ (𝑡)¿ 𝑠 ’ ’ (𝑡)

Example: Find the position, velocity, speed and acceleration at times and if

𝑣 (𝑡)=𝑠 ’ (𝑡)=3 𝑡 2 – 6 𝑡𝑎(𝑡)=𝑣 ’ (𝑡)=6 𝑡 – 6

Find :

Find :

Find

Try Find and if Find

Find

¿ 4 𝑡 3 – 6 𝑡

¿12 𝑡 2– 6

Using Anti-Derivatives with PVA

Position:

Velocity: v(t)

Acceleration: a(t)

Speed =

dttv

¿ 𝑠 ’ (𝑡)

¿𝑣 ’ (𝑡)=𝑠 ’ ’(𝑡 )

a t dt

Example: Find position, velocity, speed and acceleration at time if and where , and v(1) = 3 [so when t=1, s=2 and v=3]First find v(t)

Now find s(3), v(3), | v(3) |, a(3)

dtttv 6

213v t t c

211 3 1 3v c

1 6c 23 6v t t

Next find s(t) 23 6s t t dt 3 6s t t t c

31 1 6 1 2s c 3c

3 6 3s t t t

t s(t) v(t) |v(t)| a(t)3 -12 -21 21 -18

Try: Find position, velocity, speed and acceleration at time if and where s(1) = 3 [so when ]

2a t

2

2

2 2 2

1 1 2 34

2 4

s t t dt t t c

s cc

s t t t

t s(t) v(t) |v(t)| a(t)4 12 6 6 2

An Object moving on a linear path (like a train on train tracks)The Object is not moving (at rest) when the Velocity is zeroThe Object is moving right when the velocity is positiveThe Object is moving left when the velocity is negative

Example: A particle moves along a linear path according the position equation . Find when the particle is not moving, and the intervals when the particles moves left and right.

23 6v t t t Find v(t):Set v(t)=0, then solve for t to find where the particle is not moving:

23 6 0v t t t

3 2 0v t t t

0, 2t t

Interval test:v(t) + – +

0 2Moves left (0, 2) andright (-, 0) (2,)

are Critical #’s

t s0 22 -2 –2 0 2

To Graph:

t=2 t=0

Try: A particle moves along a linear path according the position equation s(t) = 4t3 – 12t. Find when the particle is not moving, and the intervals when the particles moves left and right.V(t) = 12t2 – 12 = 012(t + 1) (t – 1) = 0Critical #’s t = 1, t = -1 particle is not moving

Interval test:v(t) + – +

-1 1

Moves left (-1, 1) and moves right (-,-1) (1,)

Lesson VelocityDay 2: PVA with projectiles and gravity

Position Function (moon, earth, etc.) :

Feet: ft/sec2 Meters: m/sec2

• is time• is the height at time • is the acceleration due to gravity• is initial velocity (the velocity when )• is initial position (the position when )

Launched from ground level means the height s0 = 0

Example: A projectile is launched vertically upward from ground level with an initial velocity of 80 ft/sec.

A) Find the velocity at t = 1 sec and t = 3 sec.

B) How high will the projectile rise?

C) find the speed of the projectile when it hits the ground.

Step 1: write the position: Step 2: write the velocity:

ttts 8016 2 32 80v t t

Step 3: Find v(1) and v(3)

1 48 , 3 16sec secft ftv v

Step 1: Find t when 𝑡=80 /32=2.5 sec

Step 2: Find ¿−16 (2.5)2+80(2.5)=100 𝑓𝑡

Step 1: Find t when 𝑠(𝑡 )=0=−16 𝑡 2+80 𝑡=−16 𝑡 (𝑡 – 5)𝑡=0 ,𝑡=5

¿𝑣 (5)∨¿∨−32(5)+80∨¿ 80 𝑓𝑡 /sec

v0 = 80

•An object is dropped…free fall… v0 is 0

•Ground level…hits the water.. s0 is 0

• How long?...when… Find time• How long to hit the ground?... Find t when s(t)= 0• The parabola’s vertex is the maximum (or )• How long to reach max height?... Find when • Find max height… Find when then find • Speed… |v(t)|

• At impact…when it hits the ground…find t when s(t)=0

Phrases to Look for

Example: An object is dropped from a bridge and hits the water below in 3 seconds. How many feet high was the object when dropped?write the position: 0 0v 2

016s t t s

3 0s What else do you know?Hits the water at t=10 seconds =>

203 16 3 0s s

203 16 3 0s s

0 144s ft