acceleration. review distance (d) – the total ground covered by a moving object. displacement ( ...

AccelerationAcceleration

ReviewReview

Distance (d) – the total ground covered by Distance (d) – the total ground covered by a moving object.a moving object.

Displacement (Displacement (x) – the difference x) – the difference between an object’s starting position and between an object’s starting position and its ending position.its ending position.

Speed (s) – the rate at which an object is Speed (s) – the rate at which an object is moving = distance / timemoving = distance / time

Velocity (v) – the rate at which an object is Velocity (v) – the rate at which an object is displaced = displacement / timedisplaced = displacement / time

ReviewReview

Scalar quantity – a quantity that has a Scalar quantity – a quantity that has a magnitude only.magnitude only. Examples: distance and speedExamples: distance and speed

Vector quantity – a quantity that has both Vector quantity – a quantity that has both magnitude magnitude andand direction. direction. Examples: displacement and velocityExamples: displacement and velocity

Questions for ConsiderationQuestions for Consideration

What is acceleration?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 What are some useful equations involving acceleration?acceleration?

What is the acceleration due to gravity on What is the acceleration due to gravity on Earth?Earth?


AccelerationAcceleration – change in velocity over time. – change in velocity over time.

Acceleration is a vector quantity.Acceleration is a vector quantity.

Three ways an object can accelerate:Three ways an object can accelerate: speed upspeed up slow downslow down change directionschange directions

ov - vva = =

t t


Units for acceleration:Units for acceleration: a = a = v / tv / t (units for acceleration) (units for acceleration)

= (m/s) / (s)= (m/s) / (s) (units for acceleration) (units for acceleration)

= m/s/s = = m/s/s = m/sm/s22


Suppose a car accelerates from rest to Suppose a car accelerates from rest to 60.0 m/s in 12.0 s. What is the car’s 60.0 m/s in 12.0 s. What is the car’s acceleration?acceleration?

a = a = v / tv / t

a = (60.0 m/s – 0 m/s) / 12.0 sa = (60.0 m/s – 0 m/s) / 12.0 s

a = (60.0 m/s) / 12.0 sa = (60.0 m/s) / 12.0 s

a = 5.00 m/sa = 5.00 m/s22


If a car accelerates at 2.50 m/sIf a car accelerates at 2.50 m/s22, how long , how long will it take to go from 10.0 m/s to 30.0 will it take to go from 10.0 m/s to 30.0 m/s?m/s?

a = a = v / tv / t

2.50 m/s2.50 m/s22 = (30.0 m/s – 10.0 m/s) / t = (30.0 m/s – 10.0 m/s) / t

2.50 m/s2.50 m/s22 = (20.0 m/s) / t = (20.0 m/s) / t

t = (20.0 m/s) / (2.50 m/st = (20.0 m/s) / (2.50 m/s22))

t = 8.00 st = 8.00 s


A car traveling at 30.0 m/s applies the brakes A car traveling at 30.0 m/s applies the brakes and undergoes an acceleration of -3.50 m/sand undergoes an acceleration of -3.50 m/s22. . What is the car’s velocity after 3.00 seconds?What is the car’s velocity after 3.00 seconds?a = a = v / tv / t-3.50 m/s-3.50 m/s22 = = v / (3.00 s)v / (3.00 s)v = -10.5 m/sv = -10.5 m/s

v = v – vv = v – vo o

-10.5 m/s = v – 30.0 m/s-10.5 m/s = v – 30.0 m/sv = 19.5 m/s v = 19.5 m/s

Important Equations Involving Important Equations Involving AccelerationAcceleration

ov - vva = =

t t

2o

1x = v t + at2

2 2ov = v + 2a x


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

acceleration, aacceleration, a What do we already know?What do we already know?

vvoo = 0 m/s = 0 m/sv = 80.0 m/sv = 80.0 m/sx = 800. mx = 800. m


What equation will give us What equation will give us aa when when vv, , vvoo, ,

and and xx are known? are known? vv22 = v = voo

22 + 2a + 2axx (80.0 m/s)(80.0 m/s)22 = (0 m/s) = (0 m/s)22 + 2a(800. m) + 2a(800. m) 6.40x106.40x1033 m m22/s/s22 = 2a(800. m) = 2a(800. m) 8.00 m/s8.00 m/s22 = 2a = 2a a = 4.00 m/sa = 4.00 m/s22


A car is moving at 35.0 m/s when the A car is moving at 35.0 m/s when the driver slams on the brakes. If the car’s driver slams on the brakes. If the car’s acceleration is -3.00 m/sacceleration is -3.00 m/s22, how far will the , how far will the car go before coming to a stop?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, The displacement of the car while braking, xx What do we already know?What do we already know?

vvoo = 35.0 m/s = 35.0 m/s

v = 0 m/sv = 0 m/s

a = -3.00 m/sa = -3.00 m/s22


What equation allows us to calculate What equation allows us to calculate x x when when aa, , vv, and , and vvoo are known? are known? vv22 = v = voo

22 + 2a + 2axx (0 m/s)(0 m/s)22 = (35.0 m/s) = (35.0 m/s)22 + 2(-3.00 m/s + 2(-3.00 m/s22))xx 0 m0 m22/s/s22 = 1230 m = 1230 m22/s/s22 + (-6.00 m/s + (-6.00 m/s22))xx -1230 m-1230 m22/s/s22 = (-6.00 m/s = (-6.00 m/s22))xx x = 205 mx = 205 m


An airplane traveling in a straight line at An airplane traveling in a straight line at 100. m/s accelerates for 45.0 seconds at 100. m/s accelerates for 45.0 seconds at 2.00 m/s2.00 m/s22. How far does the plane travel . How far does the plane travel during this time?during this time? What do we want to know?What do we want to know?

The plane’s displacement, The plane’s displacement, x.x. What do we already know?What do we already know?

vvoo = 100. m/s = 100. m/s

a = 2.00 m/sa = 2.00 m/s22

t = 45.0 st = 45.0 s


What equation allows us to calculate What equation allows us to calculate x when x when vvoo, a, a, and , and tt are known? are known? x = vx = voot + ½ att + ½ at22

x = (100. m/s)(45.0 s) + ½ (2.00 m/sx = (100. m/s)(45.0 s) + ½ (2.00 m/s22)(45.0 s))(45.0 s)22

x = (4.50x10x = (4.50x1033 m) + ½ (2.00 m/s m) + ½ (2.00 m/s22)(2030 s)(2030 s22)) x = (4.50x10x = (4.50x1033 m) + (2030 m) m) + (2030 m) x = 6530 mx = 6530 m

Acceleration Due to GravityAcceleration Due to Gravity

Gravity accelerates objects downward.Gravity accelerates objects downward. On Earth, aOn Earth, agravitygravity = -9.81 m/s = -9.81 m/s22..

Neglecting air resistance.Neglecting air resistance. We sometimes use the letter We sometimes use the letter gg to refer specifically to to refer specifically to

accel. due to gravity.accel. due to gravity.

Acceleration Due to GravityAcceleration Due to GravityA boy drops a stone from a bridge. The A boy drops a stone from a bridge. The stone hits the water 1.50 seconds later. stone hits the water 1.50 seconds later. Calculate the height of the bridge above the Calculate the height of the bridge above the water. Ignore air resistance.water. Ignore air resistance. What do we want to know?What do we want to know?

The height of the bridge above the water.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 The displacement of the rock as it fell, or x.x. What do we already know?What do we already know?

The initial speed of the stone, vThe initial speed of the stone, voo, was 0 m/s, was 0 m/s

The acceleration is -9.81 m/sThe acceleration is -9.81 m/s22..The stone took 1.50 s to reach the water.The stone took 1.50 s to reach the water.


What equation allows us to calculate What equation allows us to calculate x x when we know when we know aa, , vvoo, and , and tt?? x = vx = voot + ½ att + ½ at22

x = (0 m/s)(1.50 s) + ½ (-9.81 m/sx = (0 m/s)(1.50 s) + ½ (-9.81 m/s22)(1.50 s))(1.50 s)22

x = ½ (-9.81 m/sx = ½ (-9.81 m/s22)(2.25 s)(2.25 s22)) x = ½ (-22.1 m)x = ½ (-22.1 m) x = -11.1 mx = -11.1 m

The stone traveled 11.1 m downward, so The stone traveled 11.1 m downward, so that’s the height of the bridge above the that’s the height of the bridge above the water.water.

acceleration. review distance (d) – the total ground covered by a moving object. displacement ( ...

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