1. Physics
Acceleration

2. Acceleration
Acceleration is defined as the rate at which velocity changes.
Acceleration is large when there is a large velocity change in a small time interval.
The ratio is called the average acceleration between the 2 times.

3. Units of acceleration Acceleration is Relative.
Velocity is measure in meters per second, m/s, so acceleration is measured in m/s/s or m/s2.
Since position and velocity are vectors, so is acceleration.
In straight-line motion, acceleration can be positive or negative.

4. Acceleration Practice Problems
The velocity of a car increases from 2.0 m/s at 1.0 s to 16 m/s at 4.5 s. What is the car’s average acceleration?
A car goes faster and faster backwards down a long driveway. We define forward velocity as positive, so backward velocity is negative. The car’s velocity changes from -2.0 m/s to -9.0 m/s in a 2.0-s time interval. Find its acceleration.

5. Answers
Given: Unknown: a v1 = 2.0 m/s Equation: v2 = 16 m/s t1 = 1.0 s t2 = 4.5 s
Solution: 14 m/s = 4.0 m/s s

6. Answers Given: Unknown: a V1 = -2.0 m/s Equation: V2 = -9.0 m/s
Dt = 2.0 s
Solution:
-9.0 m/s – (-2.0 m/s) = -7.0 m/s
A = -7.0 m/s = -3.5 m/s2
2.0 s

7. Negative Acceleration
Negative acceleration can be 1 of 2 things.
The object could be slowing down and have a negative acceleration or the object could be
speeding up but moving in a negative direction.

8. Uniform Acceleration vf = vi + at
Acceleration that does not change in time is uniform or constant acceleration.
A velocity-time graph of constant acceleration is a straight line.
Initial velocity (v1) is the velocity at time = 0s. The slope of the line is the acceleration, a, so the equation that describes the curve is:
vf = vi + at

9. Final Velocity After Uniform Acceleration
Ex: If a car with a velocity of 2.0 m/s at t = 0 accelerates at a rate of +4.0 m/s2 for 2.5 s, what is its velocity at time t = 2.5 s?
Given: vi =2.0 m/s Unknown: vf
a = 4.o m/s2
t = 2.5 s
Solution: vf = 2.0 m/s + (4.0 m/s2)(2.5 s)
= 12 m/s

10. Displacement When Velocity and Time are Known
To find the displacement if the object is uniformly accelerating, the velocity is replace by the average velocity. d = ½ (vf + vi)t
On a velocity-time graph for an object with constant acceleration, the displacement if the total area under the curve.

11. Displacement When Acceleration and Time are Known
If the initial velocity, acceleration, and time interval are known, the displacement of the object can be found by combining equations already used. d = vit + ½ at2
Example: A car starting from rest accelerates uniformly at 6.1 m/s2 for 7.0 s. How far does the car move?

12. Solution Given: Unknown: Vi = 0 m/s d = ? a = 6.1 m/s2 t = 7.0 s
d = (0 m/s)(7.0 s) + ½ (6.1 m/s2)(7.0 s)2
Answer: 150 m

13. Displacement When Velocity and Acceleration are Known
The equations for final velocity and displacement can be combined to form an equation relating initial and final velocities, acceleration, and displacement in which times does not appear.
Vf2 = vi2 + 2ad
Example: An airplane must reach a velocity of 71 m/s for takeoff. If the runway is 1.0 km long, what must the constant acceleration be?

14. Solution Given: Unknown: Vi = 0 m/s a = ? Vf = 71 m/s
d = 1 km or 1000 m
Solution: (71m/s)2 = (0 m/s)2 + 2a(1000 m)
5041 m2/s2 = 2000(a) m
(5041 m2/s2 )/(2000m) = a
a = 2.52 m/s2

15. Acceleration Due to Gravity
Galileo was the first to show that all objects fall to Earth with a constant acceleration.
No matter what the mass of the object is or what height it is dropped from, the acceleration due to gravity is always the same.
Acceleration due to gravity is called “g” and is always negative (it pulls objects downward).

16. The Value of “g”
On the surface of Earth, a freely falling object has an acceleration, g, or -9.8 m/s2.
Actually, g varies slightly from m/s2 in southern Florida to m/s2 in northern Maine.
The value of g is also smaller at high altitudes; g = m/s2 on Pike’s Peak.
The value of g would be higher at sea level.

17. Freely Falling Objects
Ex: The time the Demon Drop ride at Cedar Point, Ohio is freely falling in 1.5 s. A. What is its velocity at the end of this time? B. How far does it fall?
Given: Unknown:
g = -9.8 m/s2 vf = ?
vi = 0 m/s
t = 1.5 s

18. Solution vf = 0 + (-9.8 m/s2)(1.5 s) vf = -15 m/s
d = vit + ½ (-9.8 m/s2)(1.5 s)2
d = 0 + (-11 m)
d = -11 m
Negative displacement indicates downward motion.