2.5 Motion Equations for Constant Acceleration in One Dimension

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.