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Horizontal Kinematics

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Question 1 0:19
  • Question 2 2:19
  • Question 3 3:16
  • Question 4 4:36
  • Question 5 6:43

Transcription: Horizontal Kinematics

Hi everyone and welcome back to

In this mini-lesson, we are going to talk about horizontal kinematics and just go through page 1 of the APlusPhysics worksheet and you can find the link to that down below.0002

Note again that these are not AP level problems, but you will have that basic fundamental understanding.0012

Diving into question 1 -- You have a 747 traveling at 70 m/s to the north touching down on a runway.0018

The jet slows to rest at the rate of 2 m/s2.0026

Calculate the total distance the jet travels on the runway as it is brought to rest.0030

Let us start by writing down what we know.0036

Its start initial velocity is 70 m/s north, so V-initial is 70 m/s to the north and it touches down on the runway and it is going to slow to rest.0038

Its final velocity is going to be 0 and the acceleration is 2 m/s2, but that has to be a negative acceleration if it is slowing to rest, so that must be -2 m/s2, if we are calling north positive.0047

Find the total distance the jet travels on the runway as it is brought to rest.0064

The way I would do that is to pick a kinematic equation that would allow us to find the distance it travels and I would probably use something like V-final2 = V-initial2 + 2aδx.0070

In solving for δx, the displacement, δx will be V2 - V-initial2/2a.0085

As we substitute in our variables here, we have δx = 02 - 70 m/s2/2 × -2 m/s2 (acceleration) and that gives me at the top...0094

...702 = 4900, so -4900 m2/s2/-4 m/s2 or the total of 1,225 m.0111

All right. So that was Number 1.0135

Number 2 -- We are going to follow on with this problem and now it says beginning at point (P) -- here we have point (P) -- draw a vector to represent the magnitude and direction of the acceleration of the jet as it comes to rest.0137

Use a scale of 1 cm = 1/2 m/s2.0151

If its velocity was 70 m/s north and it is slowing down, we know that its acceleration must be to the south because it is a negative value of -2.0157

We need to draw an arrow to represent its acceleration and we need to use a scale of 1 cm, which is 1/2 a m/s2, so if we want 2 m/s2 for our acceleration, our vector must be 4 cm long.0166

I would draw something that looks kind of like that, making sure that that distance is 4 cm.0185

Number 3 -- An observer recorded the following data for the motion of a car undergoing constant acceleration.0194

What is the magnitude of the acceleration of the car?0203

We have this graph or this table of information and it is giving us time and speed.0205

Acceleration is change in velocity divided by time. And we could pick any of these changes in velocity we want.0212

We could look between those two points; we could look between those two points, or between those two points and because it is a constant acceleration, they should all be the same.0219

Let us just make this simple and go with that is our initial and that is our final point, so acceleration is change in velocity over time and δ anything is the final value minus the initial value divided by the time.0229

Our final value of speed was 7 m/s, our initial is 4, and the time it took to do that -- well from 3 to 5 seconds is going to be 2 s, so I end up with 3 m/s/2s, which is of course 1.5 m/s 2 -- Answer 3.0247

Number 4 -- We have a car traveling here on a straight road at 15 m/s, so our initial velocity is 15 m/s and it accelerates uniformly to a speed of 21 m/s, so our final velocity is 21 m/s in a time of 12 s.0277

We need to find the total distance traveled by the car in this 12 s time interval.0295

As I look at my kinematic equations, if I am looking for δx or the distance traveled, I do not see anything that jumps out at me as a way to get it directly.0300

What I am going to do here is I am going to solve for the acceleration first.0310

I know that acceleration is change in velocity over time, which will be final minus initial velocity over time or 21 m/s - 15 m/s/12 s...0315

...which is 6 m/s/12 s, therefore I would say our acceleration is going to be 0.5 m/s2.0331

I have another item that I know.0341

Now I can use any of the kinematic equations that has my δx in it in order to find the distance traveled by the car.0345

I am going to go with δx = V-initial(t) + 1/2at2...0353

...which is going to be 15 m/s (initial velocity) × 12 s (time) + 1/2 × 0.5 m/s2 (acceleration) × 12 s2 (time).0362

I put all of that into my calculator and we have 15 × 12 = 180 + 1/2 × 1/2 × 122 (144) = 216 m -- Answer 3.0379

Let us try one more. A race car starting from rest accelerates uniformly at 4.9 m/s2.0402

What is the car's speed after it has traveled 200 m?0409

Let us start with our given. It starts from rest, so that is V-initial = 0.0413

It accelerates uniformly at 4.9 m/s2, so there is our acceleration and we want to find the car's speed after it has traveled 200 m.0418

To do this one, I would use something like V2 = V-initial2 + 2aδx...0433

...which implies then that V2 = V02, so that is easy since 02 = 0.0445

That is going to be equal to 2 × 4.9 m/s2 (acceleration) × 200 m (δx), which implies then that V2 = 2 × 4.9 × 200 m...0451

...which is 1960 m2/s2, and if I take the square root of both sides, then I find that my velocity is going to be equal to about 44.3 m/s, so the correct answer is Number 3.0468

That concludes page 1 of our horizontal kinematics worksheet. Thanks so much for your time and make it a great day everyone!0489