Chapter 2: Describing Motion – Kinematics in One Dimension

Terms you should know Kinematics – how stuff moves Dynamics – why stuff moves Translational motion – moving without rotating – Like a car going down the road

It’s all relative If you are standing still on the road and a car passes you going 45mi/hr, the car moves away from you at 45mi/hr (Duh). But, if you were in a car going 35mi/hr and a car passes you going 45mi/hr, how fast does the car move away from you? It all depends on your FRAME OF REFERENCE

Displacement vs. Distance Displacement = how far you are from a certain point. Distance = how far you traveled These are not always the same thing. Example – If I drive 150 miles to Houston and 150 miles back home my displacement is zero (I’m back at home) but my distance is 300 miles.

Average Velocity When talking about how stuff moves, it is usually useful to talk about how fast it is moving. In everyday language speed and velocity to mean the same thing, but this is not quite right. Average speed = distance / time Average velocity = displacement / time

Back to my example If my whole trip to Houston and back took 4 hours of driving, then my avg. speed = distance / time = 300mi/4hr = 75mi/hr and my avg. velocity = displacement / time = 0mi / 4hr = 0 mi/hr

How to calculate displacement In order to calculate the average velocity of an object you will need to know how to calculate its displacement. Displacement = final position – initial position Note – unless it says otherwise, you can make the initial position 0. Example – If I drive 10 miles from my house then my displacement = 10mi – 0mi = 10mi

What the math looks like v bar = avg. velocity, x 2 is the final position, x 1 is the initial position, t 2 is the end time, and t 1 is the start time. Δ means “change in”

Example 1 During a 3.0s time interval, a runner’s position changes from 50.0m to 30.5m. What was the runner’s average velocity?

Solution Step 1 – given t 2 = 3s, t 1 = 0s, x 1 = 50m, and x 2 = 30.5m Step 2 – we are asked to find the average velocity. Step 3 – use v = Δx/Δt Step 4

Example 2 How far can a cyclist travel in 2.5hr in a straight line if her average velocity is 18km/hr?

Solution Step 1 – we are given v = 18km/hr and Δt = 2.5hr Step 2 – we are asked to find distance, Δx Step 3 – we can rewrite v = Δx/Δt for Δx Step 4 – Δx = v Δt = 18km/hr * 2.5hr = 45km Step 5 – check: v = Δx/Δt does 45km / 2.5hr = 18km/hr? yes, check.

Your Turn What must your car’s average velocity be to travel 280km in 3.2hr? A bird can fly 15km/hr. How long does it take to fly 75km?

Acceleration Sometimes when talking about how an object moves we want to know how fast it can change its velocity. Acceleration is the term we give to change in velocity per change in time. Average acceleration = change of velocity / time elapsed

What the math looks like Where a bar = avg. acceleration, v 2 is the final velocity, v 1 is the starting velocity, t 2 is the end time, and t 1 is the start time.

Example A car accelerates along a straight road from rest to 75km/hr in 5.0s. What is the magnitude of its average acceleration?

Solution Step 1 – we are given v 1 = 0 (it says “from rest”), v 2 = 75km/hr, and Δt = 5.0s (t 1 = 0 unless otherwise stated) Step 2 – we are asked to find the average acceleration. Step 3 – use Step 4 -

Example 2 – Deceleration A car going 15.0m/s hits the brakes and slows down to 5.0m/s over a period of 5.0s. What is the car’s average acceleration?

Solution Step 1 – we are given v 1 = 15m/s, v 2 = 5m/s and Δt = 5.0s Step 2 – we are asked to find the average acceleration. Step 3 – use Step 4 –

Your Turn A sports car accelerates from rest to 95km/hr in 6.2s. What is its average acceleration? What is it in m/s 2 ?

Motion at constant acceleration The easiest case to explore is motion where the acceleration is constant, so we will begin there. If I am accelerating steadily, then my final velocity depends on 3 things: my starting speed, my acceleration, and how long I accelerated for. v = v 0 + at

Constant acceleration cont. Likewise, distance traveled will depend on where I started, how fast I traveled, and for how long I traveled. x = x 0 + Because velocity is changing steadily, then the average velocity is just the mathematical average.

Combine ‘em We can combine our last 3 equations to get a very useful equation.

Last one I promise If we rewrite v = v 0 + at for t, we get t = (v – v 0 ) / a If we plug this in for t in x = x 0 + we get our final equation:

The 4 kinematic equations v = v 0 + at These 4 equations govern all motion (for constant acceleration)

Example You are designing an airport for small planes. One type of plane that will use the runway must reach a take off speed of 27.8 m/s and can accelerate at a rate of 2.0m/s 2. If the runway is 150m long can the plane reach take off speed? If not, what is the minimum length the runway must be?

Solution 1 Step 1: given: x 0 = 0m, v 0 = 0m/s, x = 150m, and a = 2.0m/s 2 Step 2: what is v? Step 3: need to use v 2 = v a(x – x 0 ) Step 4: v 2 = 0 + 2(2.0m/s 2 )(150m – 0m) = 600m 2 /s 2 v = √(600m 2 /s 2 ) = 24.5m/s So the runway is too short and the plane would crash!

Solution 2 Step 1: given: x 0 = 0m, v 0 = 0m/s, v = 27.8m/s, and a = 2.0m/s 2 Step 2: we are asked to find x Step 3: we need to solve v 2 = v a(x – x 0 ) for x Step 4: (x – x 0 ) = (v 2 – v 0 2 )/2a =

Example How long does it take a car to cross a 30.0m wide intersection after the light turns green, if it accelerates from rest at a rate of 2.0m/s 2 ?

Solution Step 1: given: x 0 = 0m, v 0 = 0m/s, a = 2.0m/s 2 Step 2: asked to find t Step 3: we should use Step 4: x = ½at 2 so t 2 = 2x/a so t = √(2x/a) = √(2*30m / 2.0m/s 2 ) = 5.48s

Your turn A car accelerates from 12m/s to 21m/s in 6.0s. What was its acceleration? How far did it travel in this time?

Practice makes perfect A car slows down from 25m/s to rest in a distance of 75m. What was its acceleration?

Falling Bodies The kinematic equations still apply in the vertical direction with a few tweaks. Change all x’s to y’s (for the y axis) Change a to g, g is the acceleration due to gravity and g = 9.8m/s 2

Example A ball is dropped from a tower that is 70.0m high. How far will it have fallen after 2.0s? Assume y is positive in the down direction.

Solution Step 1 – given: y 0 = 0m, v 0 = 0, t = 2s, and a = g = 9.8m/s 2 Step 2 – How far will it fall = y Step 3 – use y = y 0 + v 0 t + ½at 2 Step 4 – y = ½(9.8m/s 2 )(2s) 2 = 19.6m

The Man of Steel In the original comics, Superman could not fly, he just jumped really high. It was said his max jump was about 200m (easily the height of a “tall building” back then) What take off speed does Superman need to reach 200m?

Additional Practice A car is capable of an acceleration of 1.6m/s 2. At this rate, how long does it take to accelerate from 80km/h to 110km/hr?

Practice Problem 2 A sport’s car is advertised to be able to stop in a distance of 55m from a speed of 100km/hr. What is its acceleration in m/s 2 ? How many g’s is this?

Practice Problem 3 A light plane must reach a speed of 32m/s for take off. How long a runway is needed if the acceleration is 3.0m/s 2 ?

Practice Problem 4 A world-class sprinter can reach a top speed of about 11.5m/s in the first 15.0m of a race. What is the average acceleration of this sprinter and how long does it take her to reach that speed?

Practice Problem 5 In coming to a stop, a car leaves a skid mark 75m long on the highway. Assuming a deceleration of 7.00m/s 2, estimate the speed of the car just before breaking.

Practice Problem 6 A car traveling at 95km/hr hits a tree. The front end of the car compresses and the driver comes to a rest after traveling 0.80m. What was the average acceleration of the driver during the collision? What is this in g’s?

Practice Problem 7 A stone is dropped from the top of a cliff. It is seen to hit the ground after 2.75s. How high was the cliff?

Practice Problem 8 The best rebounders in the NBA have a vertical leap of about 120cm. A) What is their initial launch speed? B) How long are they in the air?