Acceleration – Part I
Acceleration
Acceleration measures how much an object’s speed changes over a certain time.
Objects can speed up, slow down or change direction
Acceleration can be:
A change in speed
A change in direction
A change in speed & direction
Acceleration
Acceleration can be positive, negative or zero
Negative Acceleration
Positive Acceleration
Object speeds up
Object slows down or goes the opposite direction
Zero Acceleration
Constant or no speed
Constant Acceleration
When an object’s acceleration changes at a constant rate
**for now, we will ONLY be using constant acceleration**
If an object’s acceleration is 1 m/s2, then every second it will move 1 additional meter
Acceleration
Formula for acceleration:
acceleration = change in velocity (∆v)
time
Velocity: meters per seconds (m/s)
Time: seconds (s)
Acceleration: meters per second squared (m/s2)
a = vfinal - vinitial
t
m/s/s = m/s2
Example 1
a = vfinal - vinitial
t
a = 20 - 11
4
a = 9
4
a = 2.25 m/s2
A motorcycle’s velocity at the top of the hill is 11 m/s. 4 seconds later it reaches the bottom of the hill with a velocity of 20 m/s. What is the acceleration of the motorcycle?
vi = 11 m/s
vf = 20 m/s
t = 4 s
a = ?
Example 2
a = Vfinal - Vinitial
t
- 2.9 = 0 - 13
t
t (- 2.9) = - 13
A speed skater just finished a race. After she crossed the finish line, she coasted to a complete stop. If her initial speed was 13 m/s and her acceleration was -2.9 m/s2, how long did it take her to stop?
t = - 13 / - 2.9
t = 4.5 s
vi = 13 m/s
vf = 0 m/s
a = -2.9 m/s2
t = ?
Acceleration Part II: Multiple Formulas
Example 3
a = ∆v
t
What are we given?
vi, a, t
A car with an initial velocity of 6.5 m/s accelerates at a uniform rate of 0.92 m/s2 for 3.6 s. Find the final velocity of the car during this time.
What do we need to do?
Rework the formula!
Example 3
vf = vi + at
vi = 6.5 m/s
a = 0.92 m/s2
t = 3.6 s
vf = 6.5 + (0.92)(3.6)
A car with an initial velocity of 6.5 m/s accelerates at a uniform rate of 0.92 m/s2 for 3.6 s. Find the final velocity of the car during this time.
vf = 6.5 + 3.31
vf = 9.81 m/s
Acceleration Formulas
There are four, acceleration formulas that can be used depending on your givens and unknowns.
Unknown Final Velocity
Original Equation
a = ∆v = vf – vi
t t
vf = vi + at
Unknown Displacement
∆x = vit + ½at2
Unknown Final Velocity With Unknown Time
vf2 = vi2 + 2a∆x
Example 4
An automobile with an initial velocity of 4.3 m/s accelerates uniformly at a rate of 3 m/s2.
a. Calculate the final position after 5 s.
vi = 4.3 m/s
a = 3 m/s2
t = 5 s
∆x = ?
Which formula do we use?
∆x = vit + ½at2
∆x = (4.3)(5) + ½ (3)(52)
∆x = 21.5 + ½(75)
∆x = 21.5 + 37.5
∆x = 59 m
Example 4
An automobile with an initial velocity of 4.3 m/s accelerates uniformly at a rate of 3 m/s2.
b. Find the final velocity after 5 s.
vi = 4.3 m/s
a = 3 m/s2
t = 5 s
vf = ?
Which formula do we use?
vf = vi + at
vf = 4.3 + (3)(5)
vf = 4.3 + 15
vf = 19.3 m/s
Example 5
A 737 airplane needs to reach a velocity of +67m/s to lift off the ground.
a. If the airplane starts at rest and accelerates for 18s before takeoff, what is the plane’s acceleration?
vi = 0 m/s
vf = 67 m/s
t = 18s
a = ?
Which formula do we use?
a = vf - vi
t
a = 67 – 0
18
a = 3.7 m/s2
Example 5
A 737 airplane needs to reach a velocity of +67m/s to lift off the ground.
b. How long does the runway need to be?
vi = 0 m/s
vf = 67 m/s
t = 18s
a = 3.7 m/s2
∆x = ?
Which formula do we use?
∆x = vit + ½at2
∆x = (0)(18) + ½(3.7)(182)
∆x = ½ (3.7)(324)
∆x = ½ (1,198.8)
∆x = 599.4 m