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2.5 Motion Equations for Constant Acceleration in One Dimension

- There are position, velocity, and acceleration graphs.
- The little man should be moved back and forth with the mouse.

- The simulation will move the man for you if you set the position, velocity, or acceleration.

- These kayaks racing in England are examples of moving objects that can be described with keematic equations.

- The greater the car's speed, the greater the displacement in a given time.
- We don't have an equation that relates acceleration and displacement.
- The definitions of displacement, velocity, and acceleration are already covered in this section.

- Let's make some simplifications.
- Taking the initial time to be zero is a great simplification.
- Taking means the final time on the stopwatch.

- When initial time is zero, we use the subscript 0 to represent the initial values of position and velocity.
- That is the initial position and the initial speed.
- There were no subscripts on the final values.
- That is the final time, the final position, and the final velocity.
- This gives a simpler expression for elapsed time.
- It makes the expression for displacement simpler.
- It makes the expression for change in velocity simpler.

- The assumption is that acceleration is constant.
- The assumption allows us to avoid using math.
- The average and instantaneous accelerations are the same.

- The situations we can study and the accuracy of our treatment are unaffected if we assume acceleration to be constant.
- In a lot of situations, acceleration is constant.

- In many other situations, we can accurately describe motion by assuming a constant acceleration equal to the average.
- The motion can be considered in separate parts, each of which has its own constant acceleration, when the car is speeding to top speed and then brakes to stop.

- The simple average of the initial and final velocities is reflected in the equation.
- If you increase your speed from 30 to 60 km/h, your average speed is 45 km/h.

- A jogger runs down a straight stretch of road with an average speed of 4.00 m/s.

- A sketch can be drawn.

- To find the values of,, and from the statement of the problem, we have to substitute them into the equation.

- The knowns should be identified.

- The two displacements are in the same direction.

- The equation shows the relationship between displacement, average velocity and time.
- It shows that displacement is a linear function.
- If we average 90 km/h, we will get twice as far in a given time as if we average 45 km/h.

- There is a relationship between displacement and average speed.
- An object moving twice as fast as another will move twice as far from the other object.

- The definition of acceleration can be manipulated to derive another useful equation.

- A sketch can be drawn.
- The plane is decelerating so we draw the acceleration vector in the opposite direction.

- The knowns should be identified.

- It is final velocity in this case.

- Determine which equation to use.
- The final velocity can be calculated using the equation.

- When slowing down, the final velocity is less than the initial one, but still positive.
- With jet engines, reverse thrust could be maintained long enough to stop the plane.
- This is not the case and that would be indicated by a negative final velocity.

- The airplane lands with an initial speed of 70.0 m/s and slows to a final speed of 10.0 m/s before heading for the terminal.

- The direction of the acceleration is positive and negative.

- The equation gives us insight into the relationships between time, speed, and acceleration.

- In February 2010, the Space Shuttle Endeavor blasts off from the Kennedy Space Center.

- An intercontinental missile has a larger average speed than the Space Shuttle and is more difficult to destroy in the first minute or two of flight.
- The Space Shuttle obtains a greater final velocity so that it can go around the earth rather than come back down.
- The Space Shuttle can accelerate for a longer time.

- A third equation that allows us to calculate the final position of an object that is experiencing constant acceleration can be found by combining the equations above.

- Average accelerations can be achieved by dragsters.
- The dragster should accelerate from rest at a rate of 5.56 s.

- Tony "The Sarge" Schumacher of the U.S. Army started a race with a controlled burn.

- A sketch can be drawn.

- If we take to be zero, we are asked to find displacement.
- The equation can be used if we identify, and from the statement of the problem.

- The knowns should be identified.
- It is given as 5.56 s when starting from rest.

- The standard drag racing distance is one quarter of a mile.
- The answer is reasonable.
- This is an impressive displacement, but top-notch dragsters can do a quarter mile in less time than this.

- When acceleration is not zero, displacement depends on the square of elapsed time.

- A fourth useful equation can be obtained from another equation.

- A sketch can be drawn.

- The equation relates velocities, acceleration, and displacement and no time information is required.

- The values should be identified.
- Since the dragster starts from rest, we know that.
- The average was given.

- The record for the quarter mile is 347 km/h, but even this fast speed is short of it.
- The square root has two values, one positive and one negative, so we took the positive value to indicate a velocity in the same direction as the acceleration.

- The general relationships among physical quantities can be further explored by examining the equation.

- We explore one-dimensional motion in the following examples.
- The examples give an idea of problem-solving techniques.
- The box below has easy reference to the equations.

- On dry concrete, a car can decelerate at a faster rate than on wet concrete.
- The distances needed to stop a car at 30.0 m/s are on dry concrete and on wet concrete.

- A sketch can be drawn.

- List all of the known values and identify exactly what we need to solve for in order to determine which equations are best to use.
- We will use tables to set them off in the next few examples.

- We want to solve for knowns and what they are.

- The equation will help solve the problem.

- This equation is the best because it only has one unknown.
- We know the values of the other variables.

- To solve the equation, rearrange it.

- You can enter known values.

- This part can be solved the same way as Part A.
- The only difference is the speed.

- The stopping distance is the same as in Parts A and B for dry and wet concrete.
- To answer this question, we need to calculate how far the car travels during the reaction time and then add that to the stopping time.
- It is reasonable to assume that the driver's speed is constant.

- We want to solve for knowns and what they are.

- The best equation to use is identified.

- To solve the equation, plug in the knowns.

- The total displacements in the two cases of dry and wet concrete were greater than if the driver reacted instantly.

- When dry, 64.3 m + 15.0 m is 79.3 m.

- Depending on road conditions and driver reaction time, the distance needed to stop a car varies greatly.
- This example shows the braking distances for dry and wet pavement for a car that is initially traveling at 30.0 m/s.
- The total distances traveled from the point where the driver first sees a light turn red are shown.

- The displacements found in this example seem reasonable.
- It should take more time to stop a car on wet pavement.
- Reaction time adds to the displacements.
- The general approach to solving problems is more important.
- We find an appropriate equation after we identify the knowns and quantities.
- There are many ways to solve a problem.
- The solutions presented above are the shortest and can be solved by other methods.

- A car merging into freeway traffic on a ramp.

- A sketch can be drawn.

- We are asked to solve a problem.
- We identify the known quantities in order to choose a convenient physical relationship.

- We want to solve for knowns and what they are.

- The best equation should be chosen.

- We will need to rearrange the equation.
- It will be easier to plug in the knowns first.

- The equation should be simplified.
- The units of meters are canceled when they are in the same term.
- The units of seconds can be canceled by taking the magnitude of time and the unit.

- The formula can be used to solve the problem.

- It is unreasonable for a negative value for time to mean that the event happened before the motion began.
- We can discard the solution.

- There will be two solutions if an equation has an unknown squared.
- Both solutions are reasonable in some problems, but not in others.
- For a typical freeway on-ramp, the 10.0 s answer seems reasonable.

- We can go on to many other examples and applications with the basics established.
- A general approach to problem solving that produces both correct answers and insights into physical relationships has been glimpsed in the process of developing kinematics.
- The approach that will help you succeed in this task is outlined in Problem-Solving Basics.

- SI units of meters per second squared have been used to describe some examples of acceleration.
- To get a better feel for these numbers, one can measure the stopping power of a car.
- As you approach a stop sign, apply the brakes slowly.
- The passenger should note the initial speed in miles per hour and the time taken to stop.

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