Chapter 2: Describing Movement in One Dimension, 1-D Linear Kinematics
2.5 Motion Under Constant Acceleration
Authors: William Moebs, Samuel Ling, Jeff Sanny
Adapted by: Rob Pryce, Alix Blacklin
Learning Objectives
By the end of this section, you will be able to:
- Calculate displacement of an object that is not accelerating, given initial position and velocity.
- Calculate final velocity of an accelerating object, given initial velocity, acceleration, and time.
- Calculate displacement and final position of an accelerating object, given initial position, initial velocity, time, and acceleration.
We might know that the greater the acceleration of, say, a car moving away from a stop sign, the greater the displacement in a given time. But we have not developed a specific equation that relates acceleration and displacement. In this section, we develop four convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration already covered.
Notation: t, x, v, a
First, let us examine the equations we know so far:
[latex]\bar{v} = \frac {x_f - x_i} {\Delta t}[/latex], and [latex]\bar{a} = \frac {v_f - v_i}{\Delta t}[/latex],
where [latex]\Delta t = t_f - t_i[/latex] and when [latex]t_i = 0[/latex]: [latex]\Delta t = t_f = t[/latex].
Next we make the important assumption that acceleration is constant. This assumption allows us to avoid using calculus to find instantaneous acceleration. Since acceleration is constant, the average and instantaneous accelerations are equal. That is,
[latex]\bar{a} = a = \text{constant}[/latex]
so we can use the symbol [latex]a[/latex] for acceleration at all times. Assuming acceleration to be constant does not seriously limit the situations we can study nor degrade the accuracy of our treatment. For one thing, acceleration is constant in a great number of situations. Furthermore, in many other situations we can accurately describe motion by assuming a constant acceleration equal to the average acceleration for that motion. Finally, in motions where acceleration changes drastically, such as an cyclist accelerating to top speed and then slowing down to take a corner, the motion can be considered in separate parts, each of which has its own constant acceleration.
SOLVING FOR DISPLACEMENT ([latex]\Delta x[/latex]) AND FINAL POSITION ([latex]x_f[/latex]) FROM AVERAGE VELOCITY
To get our first equation we can re-arrange the average velocity equation to solve for [latex]x_f[/latex], which gives us:
[latex]x_f = x_i + \bar{v}\Delta t[/latex]
note: when [latex]t_i = 0[/latex], this equation can also be written as [latex]x_f = x_i + \bar{v}t[/latex]
Example 2.8
Calculating Displacement: How far does the jogger run?
A jogger runs down a straight stretch of road with an average velocity of 4.00 m/s for 2.00 min. What is his final position, taking his initial position to be zero?
Strategy
Draw a sketch
The final position [latex]x_f[/latex] is given by the equation
[latex]x_f = x_i + \bar{v}\Delta t[/latex]
To find [latex]x_f[/latex], we identify the values of [latex]x_i[/latex], [latex]\bar{v}[/latex], and [latex]\Delta t[/latex] from the statement of the problem and substitute them into the equation.
Solution
- Identify the knowns. [latex]\bar{v} = \text {4.00 m/s}[/latex], [latex]\Delta t = \text {2.00 min}[/latex], and [latex]x_i = \text {0.0 m}[/latex].
- Notice that velocity is in m/s, however time is in minutes. These must be in the same units (preferably SI), so we 2.00 min to seconds:
[latex]\text {2.00min} \left(\frac {60 s}{1\text{min}}\right) = 120 \text{s}[/latex] - Enter the known values into the equation:
[latex]x_f = x_i + \bar{v}\Delta t = 0 + \left(\text{4.00 m/s}\right)\left(\text{120 s}\right) = 480m[/latex]
Discussion
Velocity and final displacement are both positive, which means they are in the same direction.
The equation [latex]x_f = x_i + \bar{v}\Delta t[/latex] gives insight into the relationship between displacement, average velocity, and time. It shows, for example, that displacement is a linear function of average velocity. (By linear function, we mean that displacement depends on [latex]\bar{v}[/latex] rather than on [latex]\bar{v}[/latex] raised to some other power, such as [latex]\bar{v}^2[/latex]. When graphed, linear functions look like straight lines with a constant slope. In fact, the equation above is simply the equation for a straight line, [latex]y = mx + b[/latex] from introductory algebra). In a running race, for example, someone will go twice as far in a given time if their average velocity is twice that of another runner.
SOLVING FOR FINAL VELOCITY ([latex]v_f[/latex]) FROM AVERAGE ACCELERATION
Similarly, we can derive a second equation by re-arranging the average acceleration equation to solve for [latex]v_f[/latex], which gives us:
[latex]v_f = v_i + \bar{a}\Delta t[/latex]
Example 2.9
Calculating Final Velocity: An Airplane Slowing Down after Landing
An airplane lands with an initial velocity of 70.0 m/s and then decelerates at 1.50 m/s2 for 40.0 s. What is its final velocity?
Strategy
Draw a sketch. We draw the acceleration vector in the direction opposite the velocity vector because the plane is decelerating.
Solution
1. Identify the knowns. [latex]v_i = \text {70.0 m/s}[/latex], [latex]\bar{a} = \text{1.50 m/s}^2[/latex], [latex]\Delta t = \text {40.0 s}[/latex]
2. Identify the unknown. In this case, it is final velocity, [latex]v_f[/latex].
3. Determine which equation to use. We can calculate the final velocity using the equation above: [latex]v_f = v_i + \bar{a}\Delta t[/latex].
4. Plug in the known values and solve.
[latex]v_f = v_i + \bar{a}\Delta t = \text {70.0 m/s} + \left(\text{-1.50 m/s}^2\right)\left(\text{40.0s}\right) = \text{10.0 m/s}[/latex]
Discussion
The final velocity is much less than the initial velocity, as desired when slowing down, but still positive. With jet engines, reverse thrust could be maintained long enough to stop the plane and start moving it backward. That would be indicated by a negative final velocity, which is not the case here.
In addition to being useful in problem solving, the equation [latex]v_f = v_i + \bar{a}\Delta t[/latex] gives us insight into the relationships among velocity, acceleration, and time. From it we can see, for example, that
- final velocity depends on how large the acceleration is and how long it lasts
- if the acceleration is zero, then the final velocity equals the initial velocity ([latex]v_f = v_i[/latex]) as expected (i.e., velocity is constant)
- if [latex]\bar{a}[/latex] is negative, then the final velocity is less than the initial velocity
(All of these observations fit our intuition, and it is always useful to examine basic equations in light of our intuition and experiences to check that they do indeed describe nature accurately.)
SOLVING FOR FINAL POSITION UNDER CONSTANT ACCELERATION (i.e. when velocity is changing and [latex]\bar{a}\neq 0[/latex])
Our next equation allows us to solve for position given initial velocity and acceleration. The equation is:
[latex]x_f = x_i + v_i\Delta t + \frac{1}{2}\bar{a}\Delta t^2[/latex]
Deriving this equation is a bit more complicated. Here are the steps if you’re interested:
First, we can start with our equation for [latex]v_f[/latex]:
[latex]v_f = v_i + \bar{a}\Delta t[/latex]
Then we can add [latex]v_i[/latex] to each side of the equation and divide by two, which gives:
[latex]\frac{v_f + v_i}{2} = v_i + \frac {1}{2}\bar{a}\Delta t[/latex]
Next, we can assume that when acceleration is constant, [latex]\bar{v}[/latex] is just the simple average of the initial and final velocities:
[latex]\bar{v} = \frac{v_i + v_f}{2}[/latex]
[For example, if you steadily increase your velocity (with constant acceleration) from 3 to 6 m/s, then your average velocity during this steady increase is 4.5 m/s. This can be checked using the above equation: [latex]\frac{3 + 6 \text{m/s}}{2} = \text {4.5 m/s}[/latex] which seems logical.]
This allows us to simply the previous equation to:
[latex]\bar{v} = v_i + \frac{1}{2}\bar{a}\Delta t[/latex]
Finally, we substitute this equation for [latex]\bar{v}[/latex] into the equation for displacement, [latex]x_f = x_i + \bar{v}\Delta t[/latex], yielding:
[latex]x_f = x_i + v_i\Delta t + \frac{1}{2}\bar{a}\Delta t^2[/latex]
Example 2.10
Calculating Displacement of an Accelerating Object: Dragsters
Dragsters can achieve average accelerations of 26.0 m/s2. Suppose such a dragster accelerates from rest at this rate for 5.56 s. How far does it travel in this time?
Draw a sketch.
We are asked to find displacement, which is [latex]x_f[/latex]. We can can take [latex]x_i[/latex] to be zero. (Think about it like the starting line of a race. It can be anywhere, but we call it 0 and measure all other positions relative to it.) We can use the equation above, once we identify [latex]v_i[/latex], [latex]\bar{a}[/latex], and [latex]\Delta t[/latex] from the statement of the problem.
Solution
1. Identify the knowns. Starting from rest means that [latex]v_i = 0[/latex], [latex]\bar{a}[/latex] is given as 26.0 m/s2 and [latex]\Delta t[/latex] is given as 5.56 s.
2. Identify the equation with the missing variable ([latex]x_f[/latex]) and plug the known values into the equation to solve for [latex]x_f[/latex]:
[latex]x_f = x_i + v_i\Delta t + \frac{1}{2}\bar{a}\Delta t^2[/latex]
Since the initial position and velocity are both zero, this simplifies to:
[latex]x_f = \frac {1}{2}\bar{a}\Delta t^2[/latex]
Substituting the identified values [latex]\bar{a}[/latex] and [latex]\Delta t[/latex] gives:
[latex]x_f = \frac {1}{2}\left(\text{26.0m/s}^2\right)\left(\text {5.56 x}\right)^2 = \text {402 m}[/latex]
Discussion
If we convert 402 m to miles, we find that the distance covered is very close to one quarter of a mile, the standard distance for drag racing. So the answer is reasonable. This is an impressive displacement in only 5.56 s, but top-notch dragsters can do a quarter mile in even less time than this. Later in the text we can learn how much force the drivers are exposed to during the race.
What else can we learn by examining the equation [latex]x_f = x_i + v_it + \frac{1}{2}\bar{a}\Delta t^2[/latex]? We see that:
- displacement depends on the square of the elapsed time when acceleration is not zero. In Example 2.10, the dragster covers only one fourth of the total distance in the first half of the elapsed time.
- if acceleration is zero, then the initial velocity equals final velocity ([latex]v_i = v_f[/latex]) and the equation becomes [latex]x_f = x_i + v_i\Delta t[/latex]
SOLVING FOR FINAL VELOCITY UNDER CONSTANT ACCELERATION (i.e. when velocity is changing and [latex]\bar{a}\neq 0[/latex])
Our fourth and last equation can be obtained from another algebraic manipulation of previous equations.
If we solve [latex]v_f = v_i + \bar{a}\Delta t[/latex] for [latex]\Delta t[/latex], we get:
[latex]\Delta t = \frac {v_f - v_i}{\bar{a}}[/latex]
Substituting this and [latex]\bar{v} = \frac {v_i + v_f}{2}[/latex] into [latex]x_f = x_i + \bar{v}\Delta t[/latex], we get our last equation:
[latex]v_f^2 = v_i^2 + 2\bar{a}\Delta x[/latex]
where [latex]\Delta x = x_f - x_i[/latex]
Example 2.11
Calculating Final Velocity: Dragsters
Calculate the final velocity of the dragster in Example 2.10 without using information about time.
Strategy
Draw a sketch.
1. Identify the known values. We know that [latex]v_i = 0[/latex], since the dragster starts from rest.
Then we note that [latex]\Delta x = \text {402 m}[/latex] (this was the answer in Example 2.10, and is simply [latex]x_f - x_i[/latex], where [latex]x_f = \text {402m}[/latex] and [latex]x_i = \text {0m}[/latex].
Finally, the average acceleration was given to be [latex]\bar{a} = \text{26.0 m/s}^2[/latex].
2. Plug the knowns into the equation [latex]v_f^2 = v_i^2 + 2\bar{a}\Delta x[/latex] and solve for [latex]v_f[/latex]
[latex]v_f^2 = 0 + 2 \left(26.0 \text{m/s}^2\right)\left(\text {402m}\right)[/latex]
Thus
[latex]v_f^2[/latex] = 20904 m2/s2
To get [latex]v_f[/latex], we take the square root:
[latex]v_f = \sqrt{20904 m^2s^2} = 144.6 \text{m/s}[/latex]
An examination of the equation [latex]v_f^2 = v_i^2 + 2\bar{a}\Delta x[/latex] can produce further insights into the general relationships among physical quantities:
- The final velocity depends on how large the acceleration is and the distance over which it acts
- For a fixed deceleration, a car that is going twice as fast doesn’t simply stop in twice the distance—it takes much further to stop. (This is why we have reduced speed zones near schools.)
Putting Equations Together
In the following examples, we further explore one-dimensional motion, but in situations requiring slightly more algebraic manipulation. The examples also give insight into problem-solving techniques. The box below provides easy reference to the equations needed.
SUMMARY OF KINEMATIC EQUATIONS
[latex]x_f = x_i + \bar{v}\Delta t[/latex]
[latex]v_f = v_i + \bar{a}\Delta t[/latex]
[latex]x_f = x_i + v_it + \frac{1}{2}\bar{a}\Delta t^2[/latex]
[latex]v_f^2 = v_i^2 + 2\bar{a}\Delta x[/latex]
Example 2.12
Calculating Displacement: How Far Does a Car Go When Coming to a Halt?
On dry concrete, a car can decelerate at a rate of 7.00 m/s2, whereas on wet concrete it can decelerate at only 5.00 m/s2. Find the distances necessary to stop a car moving at 30.0 m/s (about 110 km/h) (a) on dry concrete and (b) on wet concrete. (c) Repeat both calculations, finding the displacement from the point where the driver sees a traffic light turn red, taking into account his reaction time of 0.500 s to get his foot on the brake.
Strategy
Draw a sketch.
In order to determine which equations are best to use, we need to list all of the known values and identify exactly what we need to solve for.
Solution for (a)
1. Identify the knowns and what we want to solve for. We know that
[latex]v_i = \text {30.0m/s}[/latex];
[latex]v_f = \text {0 m/s}[/latex];
[latex]\bar{a} = \text {-7.00 m/s}^2[/latex]
([latex]\bar{a}[/latex] is negative because it is in a direction opposite to velocity).
We take [latex]x_i[/latex] to be 0.
We are looking for displacement [latex]\Delta x[/latex], or [latex]x_f - x_i[/latex].
2. Identify the equation that will help up solve the problem. The best equation to use is:
[latex]v_f^2 = v_i^2 + 2\bar{a}\Delta x[/latex]
This equation is best because it includes only one unknown, [latex]x[/latex]. We know the values of all the other variables in this equation. (There are other equations that would allow us to solve for [latex]x[/latex], but they require us to know the stopping time, [latex]\Delta t[/latex], which we do not know. We could use them but it would entail additional calculations.)
3. Rearrange the equation to solve for [latex]x[/latex].
[latex]\Delta x = \frac {v_f^2 - v_i^2}{2\bar{a}}[/latex]
4. Enter known values.
[latex]\Delta x = \frac {0^2 - \left(\text{30m/s}\right)^2}{2\left(\text{-7.0m/s}^2\right)}[/latex]
Thus,
[latex]\Delta x = \text {64.3 m}[/latex] on dry concrete.
Solution for (b)
This part can be solved in exactly the same manner as Part A. The only difference is that the deceleration is –5.00 m/s2. The result is:
[latex]x_{wet} = \text{90.0m}[/latex] on wet concrete.
Solution for (c)
Once the driver reacts, the stopping distance is the same as it is in Parts A and B for dry and wet concrete. So 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 velocity remains constant during the driver’s reaction time.
1. Identify the knowns and what we want to solve for. We know that
[latex]\bar{v} = \text{30.0m/s}[/latex];
[latex]t_{reaction} = \text {0.500 x}[/latex];
[latex]a_{reaction} = \text 0[/latex].
We take [latex]x_{i-reaction}[/latex] to be 0.
We are looking for [latex]x_{f-reaction}[/latex].
2. Identify the best equation to use.
[latex]x_f = x_i + \bar{v}\Delta t[/latex] works well because the only unknown value is [latex]x_f[/latex], which is what we want to solve for.
3. Plug in the knowns to solve the equation.
[latex]x_f = 0 + \left(\text{30.0 m/s}\right)\left(\text{0.500 x}\right) = \text {15.0 m/s}[/latex]
This means the car travels 15.0 m while the driver reacts, making the total displacements in the two cases of dry and wet concrete 15.0 m greater than if he reacted instantly.
4. Add the displacement during the reaction time to the displacement when braking.
[latex]x_{braking} + x_{reaction} = x_{total}[/latex]
- 64.3 m + 15.0 m = 79.3 m when dry
- 90.0 m + 15.0 m = 105 m when wet
Discussion
The displacements found in this example seem reasonable for stopping a fast-moving car. It should take longer to stop a car on wet rather than dry pavement. It is interesting that reaction time adds significantly to the displacements. But more important is the general approach to solving problems. We identify the knowns and the quantities to be determined and then find an appropriate equation. There is often more than one way to solve a problem. The various parts of this example can in fact be solved by other methods, but the solutions presented above are the shortest.
Example 2.13
Calculating Time: A Car Merges into Traffic
Suppose a cyclist is riding down a 200m long hill with an initial velocity is 10.0 m/s. They discover their brakes aren’t working and their bike is accelerating at a constant 2.00 m/s2. How long will it take them to get to the bottom of the hill?
Strategy
Draw a sketch.
We are asked to solve for the time [latex]\Delta t[/latex]. As before, we identify the known quantities in order to choose an equation with the unknown.
Solution
- Identify the knowns and what we want to solve for. We know that
[latex]v_i = \text {10 m/s}[/latex];
[latex]\bar{a} = \text {2.0 m/s}^2[/latex];
[latex]x_i = 0[/latex];
and [latex]x_f = 200m[/latex]
(and [latex]\Delta x = \text {200m}[/latex]. - One option would be to use the equation [latex]x_f = x_i + v_it + \frac{1}{2}\bar{a}\Delta t^2[/latex] since it has the unknown quantity, [latex]\Delta t[/latex]. However, as you simplify that equation, you’ll see it has the form of a quadratic equation ([latex]ax^2 + bx + c = 0[/latex]). If you remember it, you can solve this equation using the quadratic formula, however that is not required for this text.
- Fortunately, there is another way to solve the problem. These are explained in steps 4 and 5
- First, let’s determine the final velocity at the end of the hill using:
[latex]v_f^2 = v_i^2 + 2\bar{a}\Delta x[/latex]
This gives: [latex]v_f^2 = 10^2 + 2\left(2\text {m/s}^2\right)\left(200\text {m}\right)[/latex]
Or [latex]v_f = \sqrt{900} = 30 \text {m/s}[/latex] - Then, we can use the final velocity to solve for time using: [latex]v_f = v_i + \bar{a}\Delta t[/latex]
This gives: [latex]30 \text {m/s} = 10 + \left(2.0\text {m/s}^2\right)\left(\Delta t\right)[/latex]Or [latex]\Delta t = \frac {30 - 10}{2} = 10 \text {s}[/latex]
Discussion
Often there will be multiple ways to solve a problem in kinematics (and biomechanics); sometimes this will involve solving for intermediate variables that can be used in subsequent equations (or using more complex math)
MAKING CONNECTIONS: TAKE-HOME EXPERIMENT—BREAKING NEWS
We have been using SI units of meters per second squared to describe some examples of acceleration or deceleration of cars, runners, and trains. To achieve a better feel for these numbers, one can measure the braking deceleration of a car doing a slow (and safe) stop. Recall that, for average acceleration, [latex]\bar{a} = \frac {\Delta v}{\Delta t}[/latex]. While traveling in a car, slowly apply the brakes as you come up to a stop sign. Have a passenger note the initial speed in miles per hour and the time taken (in seconds) to stop. From this, calculate the deceleration in miles per hour per second. Convert this to meters per second squared and compare with other decelerations mentioned in this chapter. Calculate the distance traveled in braking.
Check Your Understanding
A cyclist can accelerate at a rate of 2.5m/s2 during a sprint. Has long does it take them to acceleration from 7.5 m/s to their top speed of 14.5 m/s?
Solution
To answer this, choose an equation that allows you to solve for [latex]\Delta t[/latex] given only [latex]\bar{a}[/latex], [latex]v_i \text{ and, } v_f[/latex].
[latex]v_f = v_i + \bar{a}\Delta t[/latex]
Rearrange to solve for [latex]\Delta t[/latex]
[latex]\Delta t = \frac {v_f - v_i}{\bar{a}} = \frac {14.5\text{m/s} - 7.5 \text{m/s}}{2.5 \text{m/s}^2} = 2.8 \text{s}[/latex]