1. Acceleration

2. Review Distance (d) – the total ground covered by a moving object. Displacement (  x) – the difference between an object’s starting position and its ending position. Speed (s) – the rate at which an object is moving = distance / time Velocity (v) – the rate at which an object is displaced = displacement / time

3. Review Scalar quantity – a quantity that has a magnitude only. Examples: distance and speed Examples: distance and speed Vector quantity – a quantity that has both magnitude and direction. Examples: displacement and velocity Examples: displacement and velocity

4. Questions for Consideration What is acceleration? How do we know if acceleration is happening? How do we know if acceleration is happening? What are some useful equations involving acceleration? What is the acceleration due to gravity on Earth?

5. Acceleration Acceleration – change in velocity over time. Acceleration is a vector quantity. Three ways an object can accelerate: speed up speed up slow down slow down change directions change directions

6. Acceleration Units for acceleration: a =  v / t a =  v / t (units for acceleration) = (m/s) / (s) (units for acceleration) = (m/s) / (s) (units for acceleration) = m/s/s = m/s 2 (units for acceleration) = m/s/s = m/s 2

7. Acceleration Suppose a car accelerates from rest to 60.0 m/s in 12.0 s. What is the car’s acceleration? a =  v / t a = (60.0 m/s – 0 m/s) / 12.0 s a = (60.0 m/s) / 12.0 s a = 5.00 m/s 2

8. Acceleration If a car accelerates at 2.50 m/s 2, how long will it take to go from 10.0 m/s to 30.0 m/s? a =  v / t 2.50 m/s 2 = (30.0 m/s – 10.0 m/s) / t 2.50 m/s 2 = (20.0 m/s) / t t = (20.0 m/s) / (2.50 m/s 2 ) t = 8.00 s

9. Acceleration A car traveling at 30.0 m/s applies the brakes and undergoes an acceleration of -3.50 m/s2. What is the car’s velocity after 3.00 seconds? a = v / t -3.50 m/s2 = v / (3.00 s) v = -10.5 m/s v = v – vo -10.5 m/s = v – 30.0 m/s v = 19.5 m/s

10. Important Equations Involving Acceleration

11. Acceleration An F-15 Eagle must reach a minimum speed of 80.0 m/s in order to take off. If the runway is 800. meters long, what is the minimum acceleration needed to reach take-off speed by the end of the runway if the plane starts from rest? What do we want to know? What do we want to know? acceleration, a What do we already know? What do we already know? v o = 0 m/s v = 80.0 m/s  x = 800. m

12. Acceleration What equation will give us a when v, v o, and  x are known? v 2 = v o 2 + 2a  x v 2 = v o 2 + 2a  x (80.0 m/s) 2 = (0 m/s) 2 + 2a(800. m) (80.0 m/s) 2 = (0 m/s) 2 + 2a(800. m) 6.40x10 3 m 2 /s 2 = 2a(800. m) 6.40x10 3 m 2 /s 2 = 2a(800. m) 8.00 m/s 2 = 2a 8.00 m/s 2 = 2a a = 4.00 m/s 2 a = 4.00 m/s 2

13. Acceleration A car is moving at 35.0 m/s when the driver slams on the brakes. If the car’s acceleration is -3.00 m/s 2, how far will the car go before coming to a stop? What do we want to know? What do we want to know? The displacement of the car while braking,  x What do we already know? What do we already know? v o = 35.0 m/s v = 0 m/s a = -3.00 m/s 2

14. Acceleration What equation allows us to calculate  x when a, v, and v o are known? v 2 = v o 2 + 2a  x v 2 = v o 2 + 2a  x (0 m/s) 2 = (35.0 m/s) 2 + 2(-3.00 m/s 2 )  x (0 m/s) 2 = (35.0 m/s) 2 + 2(-3.00 m/s 2 )  x 0 m 2 /s 2 = 1230 m 2 /s 2 + (-6.00 m/s 2 )  x 0 m 2 /s 2 = 1230 m 2 /s 2 + (-6.00 m/s 2 )  x -1230 m 2 /s 2 = (-6.00 m/s 2 )  x -1230 m 2 /s 2 = (-6.00 m/s 2 )  x  x = 205 m  x = 205 m

15. Acceleration An airplane traveling in a straight line at 100. m/s accelerates for 45.0 seconds at 2.00 m/s 2. How far does the plane travel during this time? What do we want to know? What do we want to know? The plane’s displacement,  x. What do we already know? What do we already know? v o = 100. m/s a = 2.00 m/s 2 t = 45.0 s

16. Acceleration What equation allows us to calculate  x when v o, a, and t are known?  x = v o t + ½ at 2  x = v o t + ½ at 2  x = (100. m/s)(45.0 s) + ½ (2.00 m/s 2 )(45.0 s) 2  x = (100. m/s)(45.0 s) + ½ (2.00 m/s 2 )(45.0 s) 2  x = (4.50x10 3 m) + ½ (2.00 m/s 2 )(2030 s 2 )  x = (4.50x10 3 m) + ½ (2.00 m/s 2 )(2030 s 2 )  x = (4.50x10 3 m) + (2030 m)  x = (4.50x10 3 m) + (2030 m)  x = 6530 m  x = 6530 m

17. Acceleration Due to Gravity Gravity accelerates objects downward. On Earth, a gravity = -9.81 m/s 2. On Earth, a gravity = -9.81 m/s 2. Neglecting air resistance. We sometimes use the letter g to refer specifically to accel. due to gravity. We sometimes use the letter g to refer specifically to accel. due to gravity.

18. Acceleration Due to Gravity A boy drops a stone from a bridge. The stone hits the water 1.50 seconds later. Calculate the height of the bridge above the water. Ignore air resistance. What do we want to know? What do we want to know? The height of the bridge above the water. Which variable does that correspond to? Which variable does that correspond to? The displacement of the rock as it fell, or  x. What do we already know? What do we already know? The initial speed of the stone, v o, was 0 m/s The acceleration is -9.81 m/s 2. The stone took 1.50 s to reach the water.

19. Acceleration Due to Gravity

20. What equation allows us to calculate  x when we know a, v o, and t?  x = v o t + ½ at 2  x = v o t + ½ at 2  x = (0 m/s)(1.50 s) + ½ (-9.81 m/s 2 )(1.50 s) 2  x = (0 m/s)(1.50 s) + ½ (-9.81 m/s 2 )(1.50 s) 2  x = ½ (-9.81 m/s 2 )(2.25 s 2 )  x = ½ (-9.81 m/s 2 )(2.25 s 2 )  x = ½ (-22.1 m)  x = ½ (-22.1 m)  x = -11.1 m  x = -11.1 m The stone traveled 11.1 m downward, so that’s the height of the bridge above the water.