2.5: Motion Equations for Constant Acceleration in One Dimension (2024)

SOLVING FOR DISPLACEMENT (Δx) AND FINAL POSITION (x) FROM AVERAGE VELOCITY WHEN ACCELERATION (a) IS CONSTANT

To get our first two new equations, we start with the definition of average velocity:

\[\displaystyle \bar{v}=\frac{Δx}{Δt}\]

Substituting the simplified notation for \(\displaystyle Δx\) and \(\displaystyle Δt\) yields

\[\displaystyle \bar{v}=\frac{x−x_0}{t}.\]

Solving for \(\displaystyle x\) yields

\[\displaystyle x=x_0+\bar{v}t,\]

where the average velocity is

\[\displaystyle \bar{v}=\frac{v_0+v}{2} \label{eq5}\]

with constant \(a\).

Equation \ref{eq5} reflects the fact that, when acceleration is constant, \(v\) is just the simple average of the initial and final velocities. For example, if you steadily increase your velocity (that is, with constant acceleration) from 30 to 60 km/h, then your average velocity during this steady increase is 45 km/h. Using the equation \(\displaystyle \bar{v}=\frac{v_0+v}{2}\) to check this, we see that

\(\displaystyle \bar{v}=\frac{v_0+v}{2}=\frac{30 km/h+60 km/h}{2}=45 km/h,\)

which seems logical.

Example \(\PageIndex{1}\): 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 is given by the equation

\(\displaystyle x=x_0+\bar{v}t\).

To find \(\displaystyle x\), we identify the values of \(\displaystyle x_0, \bar{v}\), and \(\displaystyle t\) from the statement of the problem and substitute them into the equation.

Solution

1. Identify the knowns. \(\displaystyle \bar{v}=4.00 m/s, Δt=2.00 min\), and \(\displaystyle x_0=0 m\).
2. Enter the known values into the equation.

\(\displaystyle x=x_0+\bar{v}t=0+(4.00 m/s)(120 s)=480 m\)

Discussion

Velocity and final displacement are both positive, which means they are in the same direction.

SOLVING FOR FINAL VELOCITY

We can derive another useful equation by manipulating the definition of acceleration.

\(\displaystyle a=\frac{Δv}{Δt}\)

Substituting the simplified notation for \(\displaystyle Δv\) and \(\displaystyle Δt\) gives us

\[\displaystyle a=\frac{v−v_0}{t}\] (constant a).

Solving for \(\displaystyle v\) yields

\[\displaystyle v=v_0+at\](constanta).

Example \(\PageIndex{2}\):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 \(\displaystyle 1.50 m/s^2\) 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. \(\displaystyle v_0=70.0 m/s, a=−1.50 m/s^2, t=40.0s.\)

2. Identify the unknown. In this case, it is final velocity, \(\displaystyle v_f\).

3. Determine which equation to use. We can calculate the final velocity using the equation \(\displaystyle v=v_0+at\).

4. Plug in the known values and solve.

\(\displaystyle v=v_0+at=70.0 m/s+(−1.50 m/s^2)(40.0 s)=10.0 m/s\)

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.

MAKING CONNECTIONS: REAL-WORLD CONNECTION

An intercontinental ballistic missile (ICBM) has a larger average acceleration than the Space Shuttle and achieves a greater velocity in the first minute or two of flight (actual ICBM burn times are classified—short-burn-time missiles are more difficult for an enemy to destroy). But the Space Shuttle obtains a greater final velocity, so that it can orbit the earth rather than come directly back down as an ICBM does. The Space Shuttle does this by accelerating for a longer time.

SOLVING FOR FINAL POSITION WHEN VELOCITY IS NOT CONSTANT (a≠0)

We can combine the equations above to find a third equation that allows us to calculate the final position of an object experiencing constant acceleration. We start with

\(\displaystyle v=v_0+at.\)

Adding \(\displaystyle v_0\) to each side of this equation and dividing by 2 gives

\(\displaystyle \frac{v_0+v}{2}=v_0+\frac{1}{2}at\).

Since \\frac{(v_0+v}{2}=\bar{v}\) for constant acceleration, then

\(\displaystyle \bar{v}=v_0+\frac{1}{2}at\).

Now we substitute this expression for \(\displaystyle \bar{v}\) into the equation for displacement, \(\displaystyle x=x_0+\bar{v}t\), yielding

\(\displaystyle x=x_0+v_0t+\frac{1}{2}at^2\)(constant a).

Example \(\PageIndex{3}\): Calculating Displacement of an Accelerating Object - Dragsters

Dragsters can achieve average accelerations of \(\displaystyle 26.0 m/s^2\). Suppose such a dragster accelerates from rest at this rate for 5.56 s. How far does it travel in this time?

Strategy

Draw a sketch.

We are asked to find displacement, which is \(\displaystyle x\) if we take \(\displaystyle x_0\) 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 \(\displaystyle x=x_0+v_0t+\frac{1}{2}at^2\) once we identify \(\displaystyle v_0, a,\) and \(\displaystyle t\) from the statement of the problem.

Solution

1. Identify the knowns. Starting from rest means that \(\displaystyle v_0=0, a\) is given as \(\displaystyle 26.0m/s^2\) and t is given as 5.56 s.

2. Plug the known values into the equation to solve for the unknown x:

\(\displaystyle x=x_0+v_0t+\frac{1}{2}at^2\).

Since the initial position and velocity are both zero, this simplifies to

\(\displaystyle x=\frac{1}{2}at^2\).

Substituting the identified values of a and t gives

\(\displaystyle x=\frac{1}{2}(26.0 m/s^2)(5.56 s)^2\),

yielding

\(\displaystyle x=402 m.\)

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.

SOLVING FOR FINAL VELOCITY WHEN VELOCITY IS NOT CONSTANT (a≠0)

A fourth useful equation can be obtained from another algebraic manipulation of previous equations.

If we solve \(\displaystyle v=v_0+at\) for \(\displaystyle t\), we get

\(\displaystyle t=\frac{v−v_0}{a}\).

Substituting this and \(\displaystyle \bar{v}=\frac{v_0+v}{2}\) into \(\displaystyle x=x_0+\bar{v}t\), we get

\(\displaystyle v^2=v^2_0+2a(x−x_0)\)(constant a).

Example \(\PageIndex{4}\): Calculating Final Velocity: Dragsters

Calculate the final velocity of the dragster in Example \(\PageIndex{3}\) without using information about time.

Strategy

Draw a sketch.

The equation \(\displaystyle v^2=v^2_0+2a(x−x_0)\) is ideally suited to this task because it relates velocities, acceleration, and displacement, and no time information is required.

Solution

1. Identify the known values. We know that \(\displaystyle v_0=0\), since the dragster starts from rest. Then we note that \(\displaystyle x−x_0=402 m\) (this was the answer in Example). Finally, the average acceleration was given to be \(\displaystyle a=26.0 m/s^2\).

2. Plug the knowns into the equation \(\displaystyle v^2=v^2_0+2a(x−x_0)\) and solve for \(\displaystyle v\).

\(\displaystyle v^2=0+2(26.0 m/s^2)(402 m).\)

Thus

\(\displaystyle v^2=2.09×10^4m^2/s^2.\)

To get \(\displaystyle v\), we take the square root:

\(\displaystyle v=\sqrt{2.09×10^4m^2/s^2}=145 m/s\).

Discussion

145 m/s is about 522 km/h or about 324 mi/h, but even this breakneck speed is short of the record for the quarter mile. Also, note that a square root has two values; we took the positive value to indicate a velocity in the same direction as the acceleration.

Example \(\PageIndex{5}\):Calculating Displacement: How Far Does a Car Go When Coming to a Halt?

On dry concrete, a car can decelerate at a rate of \(7.00 m/s^2\), whereas on wet concrete it can decelerate at only \(5.00 m/s^2\). Find the distances necessary to stop a car moving at 30.0 m/s (about 110 km/h)

1. on dry concrete and
2. on wet concrete.
3. 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. We shall do this explicitly in the next several examples, using tables to set them off.

Solution for (a)

1. Identify the knowns and what we want to solve for. We know that \(\displaystyle v_0=30.0 m/s; v=0; a=−7.00m/s^2\) (\(\displaystyle a\) is negative because it is in a direction opposite to velocity). We take \(\displaystyle x_0\) to be \(\displaystyle 0\). We are looking for displacement \(\displaystyle Δx\), or \(\displaystyle x−x_0\).

2. Identify the equation that will help up solve the problem. The best equation to use is

\(\displaystyle v^2=v^2_0+2a(x−x_0)\).

This equation is best because it includes only one unknown, \(\displaystyle x\). We know the values of all the other variables in this equation. (There are other equations that would allow us to solve for \(\displaystyle x\), but they require us to know the stopping time, \(\displaystyle t\), which we do not know. We could use them but it would entail additional calculations.)

3. Rearrange the equation to solve for \(\displaystyle x\).

\(\displaystyle x−x_0=\frac{v^2−v^2_0}{2a}\)

4. Enter known values.

\(\displaystyle x−0=\frac{0^2−(30.0 m/s)^2}{2(−7.00 m/s^2)}\)

Thus,

\(\displaystyle x=64.3 m\) 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 \(\displaystyle –5.00 m/s^2\). The result is

\(\displaystyle x_{wet}=90.0 m\) 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 \(\displaystyle \bar{v}=30.0 m/s; t_{reaction}=0.500s; a_{reaction}=0\). We take \(\displaystyle x_{0−reaction}\) to be 0. We are looking for \(\displaystyle x_{reaction}\).

2. Identify the best equation to use.

\(\displaystyle x=x_0+\bar{v}t\) works well because the only unknown value is \(\displaystyle x\), which is what we want to solve for.

3. Plug in the knowns to solve the equation.

\(\displaystyle x=0+(30.0 m/s)(0.500 s)=15.0 m.\)

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.

\(\displaystyle x_{braking}+x_{reaction}=x_{total}\)

a. 64.3 m + 15.0 m = 79.3 m when dry
b. 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 \(\PageIndex{5}\): Calculating Time - A Car Merges into Traffic

Suppose a car merges into freeway traffic on a 200-m-long ramp. If its initial velocity is 10.0 m/s and it accelerates at \(\displaystyle .00 m/s^2\), how long does it take to travel the 200 m up the ramp? (Such information might be useful to a traffic engineer.)

Strategy

Draw a sketch.

Solution

1. Identify the knowns and what we want to solve for. We know that \(\displaystyle v_0=10 m/s; a=2.00 m/s^2\); and \(\displaystyle x=200 m.\)

2. We need to solve for t. Choose the best equation. \(\displaystyle x=x_0+v_0t+\frac{1}{2}at^2\) works best because the only unknown in the equation is the variable t for which we need to solve.

3. We will need to rearrange the equation to solve for \(t\). In this case, it will be easier to plug in the knowns first.

\(\displaystyle 200 m=0 m+(10.0 m/s)t+\frac{1}{2}(2.00 m/s^2)t^2\)

4. Simplify the equation. The units of meters (m) cancel because they are in each term. We can get the units of seconds (s) to cancel by taking \(\displaystyle t=ts\), where \(\displaystyle t\) is the magnitude of time and s is the unit. Doing so leaves

\(\displaystyle 200=10t+t^2.\)

5. Use the quadratic formula to solve for \(\displaystyle t\).

(a) Rearrange the equation to get 0 on one side of the equation.

\(\displaystyle t^2+10t−200=0\)

This is a quadratic equation of the form

\(\displaystyle at^2+bt+c=0,\)

where the constants are \(\displaystyle a=1.00,b=10.0\),and \(\displaystyle c=−200.\)

(b) Its solutions are given by the quadratic formula:

\(\displaystyle t=\frac{−b±\sqrt{b^2−4ac}}{2a}\).

This yields two solutions for \(\displaystyle t\), which are

\(\displaystyle t=10.0\) and \(\displaystyle −20.0\).

In this case, then, the time is \(\displaystyle t=t\) in seconds, or

\(\displaystyle t=10.0s\) and \(\displaystyle −20.0s\).

A negative value for time is unreasonable, since it would mean that the event happened 20 s before the motion began. We can discard that solution. Thus,

\(\displaystyle t=10.0s.\)

Discussion

Whenever an equation contains an unknown squared, there will be two solutions. In some problems both solutions are meaningful, but in others, such as the above, only one solution is reasonable. The 10.0 s answer seems reasonable for a typical freeway on-ramp.

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, \(\displaystyle \bar{a}=Δv/Δt\). 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.

Exercise \(\PageIndex{1}\)

A manned rocket accelerates at a rate of \(\displaystyle 20 m/s^2\) during launch. How long does it take the rocket to reach a velocity of 400 m/s?

Answer

To answer this, choose an equation that allows you to solve for time \(\displaystyle t\), given only \(\displaystyle a, v_0\), and \(\displaystyle v\).

\(\displaystyle v=v_0+at\)

Rearrange to solve for \(\displaystyle t\).

\(\displaystyle t=\frac{v−v_0}{a}=\frac{400 m/s−0 m/s}{20 m/s^2}=20 s\)