Can an Object With Constant Acceleration Reverse Its Direction Twice

Learning Objectives

By the stop of this section, yous volition exist able to:

• Identify which equations of motion are to be used to solve for unknowns.
• Use appropriate equations of motion to solve a 2-body pursuit problem.

Y'all might guess that the greater the acceleration of, say, a auto moving away from a end sign, the greater the car's displacement in a given time. But, we take not adult a specific equation that relates acceleration and deportation. In this section, nosotros await at some user-friendly equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration. We beginning investigate a single object in movement, chosen unmarried-body motion. Then nosotros investigate the motion of two objects, called two-trunk pursuit problems.

Notation

Commencement, let united states of america make some simplifications in notation. Taking the initial fourth dimension to be zero, as if fourth dimension is measured with a stopwatch, is a great simplification. Since elapsed time is $\text{Δ}t=t_{\text{f}}-t_0$ , taking $t_0=0$ ways that $\text{Δ}t=t_{\text{f}}$ , the terminal fourth dimension on the stopwatch. When initial time is taken to be aught, we employ the subscript 0 to denote initial values of position and velocity. That is, $x_0$ is the initial position and $v_0$ is the initial velocity. We put no subscripts on the final values. That is, t is the final fourth dimension, x is the last position, and v is the final velocity. This gives a simpler expression for elapsed time, $\text{Δ}t=t$ . Information technology also simplifies the expression for x deportation, which is now $\text{Δ}\mathrm{ten}=x-x_0$ . Also, it simplifies the expression for change in velocity, which is now $\text{Δ}v=v-5_0$ . To summarize, using the simplified notation, with the initial time taken to exist zippo,

$$\begin{array}{c}\text{Δ}t=t\hfill \\ \text{Δ}x=x-x_0\hfill \\ \text{Δ}v=v-v_0,\hfill \end{array}$$

where the subscript 0 denotes an initial value and the absence of a subscript denotes a final value in whatever motion is nether consideration.

Nosotros now brand the important assumption that acceleration is constant. This assumption allows us to avert using calculus to find instantaneous acceleration. Since acceleration is constant, the average and instantaneous accelerations are equal—that is,

$$\stackrel{\text{–}}{a}=a=\text{constant}\text{.}$$

Thus, we can utilize the symbol a for acceleration at all times. Assuming acceleration to exist constant does not seriously limit the situations we can written report nor does information technology degrade the accuracy of our treatment. For one affair, acceleration is constant in a slap-up number of situations. Furthermore, in many other situations nosotros can describe motion accurately by assuming a abiding dispatch equal to the average acceleration for that move. Lastly, for motion during which acceleration changes drastically, such as a car accelerating to top speed and and so braking to a terminate, move tin can be considered in separate parts, each of which has its ain constant acceleration.

Deportation and Position from Velocity

To get our first two equations, nosotros beginning with the definition of average velocity:

$$\stackrel{\text{–}}{v}=\frac{\text{Δ}x}{\text{Δ}t}.$$

Substituting the simplified notation for $\text{Δ}x$ and $\text{Δ}t$ yields

$$\stackrel{\text{–}}{v}=\frac{x-x_0}{t}.$$

Solving for 10 gives us

where the average velocity is

The equation $\stackrel{\text{–}}{v}=\frac{v_0+5}{\mathrm{two}}$ reflects the fact that when acceleration is constant, $\stackrel{\text{–}}{\mathrm{five}}$ is just the simple average of the initial and final velocities. Effigy 3.18 illustrates this concept graphically. In part (a) of the figure, dispatch is constant, with velocity increasing at a constant rate. The average velocity during the ane-h interval from 40 km/h to 80 km/h is 60 km/h:

$$\stackrel{\text{–}}{v}=\frac{v_0+v}{2}=\frac{40\text{km/h}+80\text{km/h}}{2}=\mathrm{threescore}\text{km/h}\text{.}$$

In part (b), acceleration is not constant. During the 1-h interval, velocity is closer to 80 km/h than 40 km/h. Thus, the average velocity is greater than in role (a).

Figure three.xviii (a) Velocity-versus-time graph with constant acceleration showing the initial and final velocities $v_0\text{and}v$. The average velocity is $\frac{\mathrm{one}}{2}(v_0+v)=60\text{km}\text{/}\text{h}$. (b) Velocity-versus-time graph with an dispatch that changes with time. The average velocity is not given past $\frac{\mathrm{one}}{\mathrm{two}}(5_0+v)$, simply is greater than 60 km/h.

Solving for Final Velocity from Acceleration and Time

We can derive some other useful equation by manipulating the definition of acceleration:

$$a=\frac{\text{Δ}v}{\text{Δ}t}.$$

Substituting the simplified notation for $\text{Δ}v$ and $\text{Δ}t$ gives usa

$$a=\frac{v-{\mathrm{five}}_0}{t}\left(\text{constant}a\right).$$

Solving for five yields

$$5={\mathrm{five}}_0+at\left(\text{constant}a\right).$$

iii.12

Example 3.7

Calculating Final Velocity

An plane lands with an initial velocity of 70.0 m/s and then accelerates opposite to the motion at ane.50 m/s² for 40.0 s. What is its concluding velocity?

Strategy

Start, we identify the knowns: $v_0=\mathrm{lxx}\text{yard/southward,}a=\mathrm{-1.50}{\text{m/s}}^2,t=40\text{s}$ .

Second, we identify the unknown; in this case, it is terminal velocity $5_{\text{f}}$ .

Concluding, we determine which equation to use. To do this nosotros effigy out which kinematic equation gives the unknown in terms of the knowns. We calculate the concluding velocity using Equation iii.12, $\mathrm{five}=v_0+at$ .

Solution

Substitute the known values and solve:

$$\mathrm{five}=5_0+at=\mathrm{lxx.0}\text{m/due south}+\left(\mathrm{-1.50}\text{m/}{\text{s}}^2\right)\left(\mathrm{40.0\; s}\right)=\mathrm{ten.0\; chiliad/s.}$$

Figure iii.19 is a sketch that shows the acceleration and velocity vectors.

Figure 3.nineteen The airplane lands with an initial velocity of seventy.0 m/s and slows to a final velocity of ten.0 m/due south earlier heading for the terminal. Notation the acceleration is negative considering its direction is opposite to its velocity, which is positive.

Significance

The final velocity is much less than the initial velocity, as desired when slowing down, only is still positive (see figure). With jet engines, reverse thrust tin can be maintained long enough to stop the plane and start moving it backward, which is indicated by a negative terminal velocity, but is non the case here.

In addition to being useful in problem solving, the equation $\mathrm{five}=v_0+at$ gives us insight into the relationships amongst velocity, dispatch, and fourth dimension. Nosotros can encounter, for example, that

• Final velocity depends on how large the dispatch is and how long it lasts
• If the acceleration is zero, then the final velocity equals the initial velocity (v = v ₀), as expected (in other words, velocity is constant)
• If a is negative, and then the last velocity is less than the initial velocity

All these observations fit our intuition. Notation that it is always useful to examine basic equations in light of our intuition and experience to cheque that they exercise indeed describe nature accurately.

Solving for Concluding Position with Constant Acceleration

Nosotros tin combine the previous equations to find a tertiary equation that allows united states to calculate the final position of an object experiencing abiding acceleration. We start with

$$v=v_0+at.$$

Adding $v_0$ to each side of this equation and dividing past 2 gives

$$\frac{{\mathrm{five}}_0+v}{2}=v_0+\frac{1}{\mathrm{two}}at.$$

Since $\frac{v_0+v}{\mathrm{ii}}=\stackrel{\text{–}}{\mathrm{five}}$ for constant dispatch, we take

$$\stackrel{\text{–}}{v}=v_0+\frac{\mathrm{i}}{2}at.$$

Now we substitute this expression for $\stackrel{\text{–}}{5}$ into the equation for deportation, $x={\mathrm{10}}_0+\stackrel{\text{–}}{v}t$ , yielding

$$x={\mathrm{ten}}_0+v_{0}t+\frac{1}{2}a{t}^{\mathrm{two}}\left(\text{constant}a\right).$$

iii.xiii

Example 3.8

Calculating Deportation of an Accelerating Object

Dragsters can achieve an average acceleration of 26.0 m/southward^two. Suppose a dragster accelerates from rest at this rate for 5.56 s Effigy 3.20. How far does it travel in this time?

Effigy 3.20 U.S. Army Top Fuel pilot Tony "The Sarge" Schumacher begins a race with a controlled burnout. (credit: Lt. Col. William Thurmond. Photograph Courtesy of U.Due south. Army.)

Strategy

First, let's draw a sketch Figure three.21. We are asked to detect displacement, which is x if we take ${\mathrm{10}}_0$ to be nil. (Remember nigh ${\mathrm{10}}_0$ as the starting line of a race. Information technology can exist anywhere, merely we call it nada and measure all other positions relative to it.) We can utilize the equation $x=x_0+{\mathrm{five}}_{0}t+\frac{\mathrm{i}}{\mathrm{ii}}a{t}^{\mathrm{ii}}$ when we place ${\mathrm{five}}_0$ , $a$ , and t from the argument of the problem.

Figure three.21 Sketch of an accelerating dragster.

Solution

First, we demand to place the knowns. Starting from rest means that $v_0=0$ , a is given equally 26.0 one thousand/southward^two and t is given as 5.56 s.

Second, we substitute the known values into the equation to solve for the unknown:

$$x={\mathrm{ten}}_0+v_{0}t+\frac{1}{2}a{t}^2.$$

Since the initial position and velocity are both nothing, this equation simplifies to

$$x=\frac{1}{2}a{t}^{\mathrm{ii}}.$$

Substituting the identified values of a and t gives

$$\mathrm{10}=\frac{1}{2}(26.0{\text{grand/south}}^2){(5.56\text{s})}^2=402\text{1000}\text{.}$$

Significance

If nosotros convert 402 m to miles, we find that the distance covered is very shut to one-quarter of a mile, the standard altitude for drag racing. And then, our reply is reasonable. This is an impressive displacement to embrace in merely 5.56 southward, but top-notch dragsters can practise a quarter mile in even less time than this. If the dragster were given an initial velocity, this would add some other term to the distance equation. If the aforementioned acceleration and fourth dimension are used in the equation, the distance covered would be much greater.

What else can we learn past examining the equation $\mathrm{10}={\mathrm{10}}_0+5_{0}t+\frac{\mathrm{ane}}{2}a{t}^2?$ We can see the following relationships:

• Deportation depends on the square of the elapsed time when acceleration is not nothing. In Instance three.8, the dragster covers but i-fourth of the full distance in the outset half of the elapsed time.
• If acceleration is zippo, and so initial velocity equals average velocity $({\mathrm{five}}_0=\stackrel{\text{–}}{v})$, and $\mathrm{ten}=x_0+{\mathrm{five}}_{0}t+\frac{1}{\mathrm{two}}a{t}^2\text{becomes}\mathrm{ten}=x_0+{\mathrm{five}}_{0}t.$

Solving for Final Velocity from Altitude and Acceleration

A fourth useful equation tin be obtained from another algebraic manipulation of previous equations. If nosotros solve $v=5_0+at$ for t, we get

$$t=\frac{\mathrm{five}-v_0}{a}.$$

Substituting this and $\stackrel{\text{–}}{v}=\frac{{\mathrm{five}}_0+v}{2}$ into $x=x_0+\stackrel{\text{–}}{v}t$ , we get

$$v^{\mathrm{ii}}=v_0^{\mathrm{ii}}+2a(x-x_0)(\text{constant}a).$$

3.fourteen

Case 3.nine

Calculating Final Velocity

Calculate the final velocity of the dragster in Example 3.8 without using information about fourth dimension.

Strategy

The equation $5^2=v_0^2+\mathrm{two}a(x-{\mathrm{ten}}_0)$ is ideally suited to this job because it relates velocities, acceleration, and deportation, and no time information is required.

Solution

First, we identify the known values. We know that v ₀ = 0, since the dragster starts from balance. We besides know that 10 − 10 ₀ = 402 thou (this was the respond in Case three.8). The average acceleration was given by a = 26.0 m/due south^two.

Second, we substitute the knowns into the equation $5^2=v_0^{\mathrm{ii}}+2a(x-x_0)$ and solve for v:

$$v^{\mathrm{two}}=0+2\left(26.0{\text{m/s}}^2\right)\left(402\text{m}\right).$$

Thus,

$$\begin{array}{ccc}{v}^2\hfill & =\hfill & \mathrm{two.09}\times {10}^4{\text{m}}^2{\text{/s}}^2\hfill \\ \\ \mathrm{five}\hfill & =\hfill & \sqrt{2.09\times {\mathrm{ten}}^4{\text{thousand}}^2{\text{/s}}^2}=145\text{m/s}\text{.}\hfill \end{array}$$

Significance

A velocity of 145 m/south is most 522 km/h, or nearly 324 mi/h, but even this breakneck speed is brusk of the record for the quarter mile. Also, note that a square root has two values; nosotros took the positive value to signal a velocity in the same management equally the acceleration.

An examination of the equation $v^2=5_0^2+2a(x-{\mathrm{10}}_0)$ can produce additional insights into the general relationships among physical quantities:

• The final velocity depends on how big the acceleration is and the distance over which it acts.
• For a stock-still acceleration, a machine that is going twice as fast doesn't simply stop in twice the distance. It takes much farther to stop. (This is why we have reduced speed zones about schools.)

Putting Equations Together

In the post-obit examples, we go on to explore i-dimensional movement, but in situations requiring slightly more algebraic manipulation. The examples also give insight into problem-solving techniques. The notation that follows is provided for like shooting fish in a barrel reference to the equations needed. Exist aware that these equations are not independent. In many situations we have two unknowns and demand two equations from the fix to solve for the unknowns. We demand equally many equations every bit there are unknowns to solve a given situation.

Summary of Kinematic Equations (constant a)

$$x=x_0+\stackrel{\text{–}}{v}t$$

$$\stackrel{\text{–}}{v}=\frac{{\mathrm{five}}_0+v}{2}$$

$$v={\mathrm{five}}_0+at$$

$$\mathrm{10}=x_0+{\mathrm{five}}_{0}t+\frac{1}{2}a{t}^2$$

$$v^2=v_0^{\mathrm{ii}}+2a\left(x-x_0\right)$$

Before we get into the examples, permit's look at some of the equations more than closely to see the behavior of acceleration at extreme values. Rearranging Equation 3.12, we accept

$$a=\frac{v-v_0}{t}.$$

From this nosotros see that, for a finite time, if the difference between the initial and final velocities is small, the dispatch is small, budgeted zero in the limit that the initial and final velocities are equal. On the contrary, in the limit $t\to 0$ for a finite departure between the initial and last velocities, acceleration becomes infinite.

Similarly, rearranging Equation 3.14, we tin limited dispatch in terms of velocities and displacement:

$$a=\frac{5^2-v_0^2}{2(\mathrm{10}-{\mathrm{10}}_0)}.$$

Thus, for a finite difference betwixt the initial and final velocities acceleration becomes infinite in the limit the displacement approaches zero. Acceleration approaches zero in the limit the difference in initial and final velocities approaches zero for a finite deportation.

Example 3.10

How Far Does a Car Go?

On dry physical, a auto can accelerate opposite to the motion at a rate of 7.00 m/southⁱⁱ, whereas on wet physical information technology tin accelerate contrary to the motility at only 5.00 k/s². Find the distances necessary to stop a car moving at 30.0 yard/s (about 110 km/h) on (a) dry concrete and (b) wet concrete. (c) Repeat both calculations and find the displacement from the betoken where the commuter sees a traffic lite turn reddish, taking into account his reaction time of 0.500 s to become his foot on the brake.

Strategy

First, we demand to draw a sketch Figure 3.22. To determine which equations are best to utilize, we need to list all the known values and identify exactly what we need to solve for.

Effigy 3.22 Sample sketch to visualize acceleration opposite to the motion and stopping distance of a car.

Solution

1. First, we need to identify the knowns and what nosotros want to solve for. We know that v ₀ = thirty.0 m/s, v = 0, and a = −7.00 chiliad/s² (a is negative considering it is in a direction opposite to velocity). Nosotros have ten ₀ to be nada. We are looking for deportation $\text{Δ}\mathrm{ten}$, or x − 10 ₀.
   2d, we place the equation that will help us solve the problem. The all-time equation to utilize is

   $$v^{\mathrm{two}}=v_0^2+2a(x-x_0).$$

   This equation is all-time considering it includes only i unknown, ten. We know the values of all the other variables in this equation. (Other equations would allow u.s.a. to solve for 10, but they crave u.s. to know the stopping time, t, which we do not know. We could use them, but it would entail additional calculations.)
   Tertiary, we rearrange the equation to solve for x:

   $$x-x_0=\frac{v^{\mathrm{two}}-v_0^{\mathrm{ii}}}{2a}$$

   and substitute the known values:

   $$x-0=\frac{0^2-{(30.0\text{m/s})}^{\mathrm{two}}}{2(\mathrm{-7.00}{\text{m/due south}}^2)}.$$

   Thus,

   $$x=\mathrm{64.iii}\text{m on dry physical}\text{.}$$

2. This part tin can be solved in exactly the same way as (a). The but departure is that the acceleration is −5.00 m/s². The result is

   $${\mathrm{ten}}_{\text{wet}}=90.0\text{thou on wet concrete.}$$

3. When the driver reacts, the stopping distance is the same as information technology is in (a) and (b) for dry out and wet concrete. So, to answer this question, nosotros need to calculate how far the automobile travels during the reaction time, and then add that to the stopping fourth dimension. It is reasonable to presume the velocity remains constant during the driver'due south reaction time.
   To exercise this, we, over again, identify the knowns and what we desire to solve for. We know that $\stackrel{\text{–}}{v}=30.0\text{m/s}$, $t_{\text{reaction}}=0.500\text{s}$, and $a_{\text{reaction}}=0$. We take ${\mathrm{10}}_{\text{0-reaction}}$ to exist aught. Nosotros are looking for $x_{\text{reaction}}$.
   Second, every bit earlier, we identify the best equation to utilize. In this case, $x={\mathrm{ten}}_0+\stackrel{\text{–}}{v}t$ works well considering the but unknown value is ten, which is what we want to solve for.
   Third, nosotros substitute the knowns to solve the equation:

   $$x=0+\left(\mathrm{three}0.0\text{m/southward}\right)\left(0.500\text{south}\right)=15.0\text{grand}.$$

   This ways the automobile travels 15.0 one thousand while the commuter reacts, making the total displacements in the two cases of dry and moisture concrete xv.0 1000 greater than if he reacted instantly.
   Terminal, we and so add the deportation during the reaction time to the displacement when braking (Figure 3.23),

   $$x_{\text{braking}}+x_{\text{reaction}}=x_{\text{total}},$$

   and find (a) to be 64.3 m + 15.0 1000 = 79.3 m when dry out and (b) to be 90.0 m + 15.0 g = 105 thousand when moisture.

Effigy three.23 The altitude necessary to cease a car varies profoundly, depending on road weather condition and driver reaction time. Shown hither are the braking distances for dry and wet pavement, every bit calculated in this case, for a motorcar traveling initially at xxx.0 m/due south. Also shown are the total distances traveled from the point when the commuter start sees a light turn red, assuming a 0.500-southward reaction fourth dimension.

Significance

The displacements found in this case seem reasonable for stopping a fast-moving auto. It should take longer to finish a car on wet pavement than dry. It is interesting that reaction time adds significantly to the displacements, but more than important is the general arroyo to solving issues. We identify the knowns and the quantities to be determined, then find an appropriate equation. If there is more than one unknown, we demand as many independent equations as at that place are unknowns to solve. There is often more than 1 way to solve a trouble. The various parts of this instance can, in fact, be solved by other methods, but the solutions presented here are the shortest.

Example 3.11

Calculating Fourth dimension

Suppose a machine merges into thruway traffic on a 200-m-long ramp. If its initial velocity is 10.0 thousand/s and it accelerates at 2.00 m/s², how long does information technology have the car to travel the 200 m upward the ramp? (Such data might be useful to a traffic engineer.)

Strategy

First, we depict a sketch Figure 3.24. We are asked to solve for time t. Every bit earlier, we place the known quantities to choose a convenient concrete human relationship (that is, an equation with one unknown, t.)

Figure 3.24 Sketch of a machine accelerating on a superhighway ramp.

Solution

Once more, we identify the knowns and what we want to solve for. We know that $x_0=0,$
$5_0=10\text{m/s},a=2.00\text{yard/}{\text{south}}^{\mathrm{two}}$ , and x = 200 thousand.

We need to solve for t. The equation $x={\mathrm{10}}_0+v_{0}t+\frac{1}{\mathrm{two}}a{t}^2$ works best because the only unknown in the equation is the variable t, for which nosotros demand to solve. From this insight we see that when nosotros input the knowns into the equation, we end upwardly with a quadratic equation.

Nosotros demand to rearrange the equation to solve for t, and so substituting the knowns into the equation:

$$200\text{g}=0\text{one thousand}+(\mathrm{ten.0}\text{thou/due south})t+\frac{\mathrm{i}}{\mathrm{ii}}\left(\mathrm{ii.00}{\text{m/s}}^{\mathrm{two}}\right)t^{\mathrm{ii}}.$$

We then simplify the equation. The units of meters cancel considering they are in each term. We can get the units of seconds to abolish by taking t = t s, where t is the magnitude of time and due south is the unit. Doing so leaves

$$200=\mathrm{ten}t+t^2.$$

We then use the quadratic formula to solve for t,

$$\begin{array}{c}{t}^{\mathrm{two}}+\mathrm{ten}t-200=0\\ \\ \\ t=\frac{\text{−}b\pm \sqrt{b^2-4ac}}{2a},\end{array}$$

which yields two solutions: t = 10.0 and t = −20.0. A negative value for fourth dimension is unreasonable, since information technology would mean the event happened 20 due south before the motion began. We tin can discard that solution. Thus,

$$t=\mathrm{x.0}\text{s}\text{.}$$

Significance

Whenever an equation contains an unknown squared, at that place are two solutions. In some problems both solutions are meaningful; in others, only one solution is reasonable. The 10.0-southward answer seems reasonable for a typical expressway on-ramp.

Cheque Your Understanding three.5

A rocket accelerates at a rate of xx one thousand/s² during launch. How long does it take the rocket to reach a velocity of 400 grand/s?

Instance three.12

Acceleration of a Spaceship

A spaceship has left Earth's orbit and is on its fashion to the Moon. It accelerates at xx m/s² for 2 min and covers a distance of 1000 km. What are the initial and last velocities of the spaceship?

Strategy

Nosotros are asked to notice the initial and last velocities of the spaceship. Looking at the kinematic equations, we run across that one equation volition not give the reply. Nosotros must utilise one kinematic equation to solve for i of the velocities and substitute information technology into some other kinematic equation to become the second velocity. Thus, we solve two of the kinematic equations simultaneously.

Solution

Outset we solve for $5_0$ using $x=x_0+{\mathrm{five}}_{0}t+\frac{\mathrm{i}}{2}a{t}^2:$

$$x-x_0=5_{0}t+\frac{1}{2}a{t}^{\mathrm{two}}$$

$$1.0\times {10}^{\mathrm{six}}\text{m}=v_0(120.0\text{s})+\frac{1}{\mathrm{ii}}(20.0{\text{g/due south}}^2){(120.0\text{s})}^{\mathrm{two}}$$

$$v_0=\mathrm{7133.iii}\text{1000/s}\text{.}$$

Then we substitute $5_0$ into $v=v_0+at$ to solve for the last velocity:

$$v={\mathrm{five}}_0+at=\mathrm{7133.iii}\text{m/southward}+(\mathrm{xx.0}{\text{yard/s}}^2)(120.0\text{south})=\mathrm{9533.three}\text{m/s.}$$

Significance

There are six variables in displacement, fourth dimension, velocity, and dispatch that describe movement in i dimension. The initial conditions of a given problem tin can be many combinations of these variables. Because of this diversity, solutions may non exist every bit easy equally simple substitutions into one of the equations. This example illustrates that solutions to kinematics may require solving ii simultaneous kinematic equations.

With the basics of kinematics established, we can go along to many other interesting examples and applications. In the process of developing kinematics, we have also glimpsed a general approach to problem solving that produces both correct answers and insights into concrete relationships. The adjacent level of complexity in our kinematics problems involves the motion of two interrelated bodies, called 2-trunk pursuit bug.

Two-Body Pursuit Problems

Up until this point we have looked at examples of movement involving a single torso. Even for the trouble with two cars and the stopping distances on wet and dry roads, we divided this problem into two separate issues to find the answers. In a 2-body pursuit trouble, the motions of the objects are coupled—meaning, the unknown nosotros seek depends on the move of both objects. To solve these problems we write the equations of motion for each object and and then solve them simultaneously to discover the unknown. This is illustrated in Figure three.25.

Figure 3.25 A ii-body pursuit scenario where automobile 2 has a constant velocity and car i is behind with a abiding acceleration. Motorcar i catches up with car 2 at a later time.

The time and distance required for car 1 to take hold of car 2 depends on the initial altitude car 1 is from motorcar 2 every bit well as the velocities of both cars and the dispatch of automobile 1. The kinematic equations describing the motion of both cars must be solved to detect these unknowns.

Consider the following instance.

Instance iii.13

Cheetah Catching a Gazelle

A cheetah waits in hiding backside a bush. The cheetah spots a gazelle running by at x m/south. At the instant the gazelle passes the cheetah, the cheetah accelerates from residuum at 4 thou/southward² to catch the gazelle. (a) How long does it take the cheetah to catch the gazelle? (b) What is the displacement of the gazelle and chetah?

Strategy

We utilize the ready of equations for abiding acceleration to solve this trouble. Since in that location are ii objects in motion, we have separate equations of move describing each animate being. But what links the equations is a common parameter that has the same value for each animal. If we await at the problem closely, it is clear the common parameter to each animal is their position x at a later time t. Since they both start at ${\mathrm{10}}_0=0$ , their displacements are the same at a after fourth dimension t, when the chetah catches upward with the gazelle. If we selection the equation of motion that solves for the displacement for each animal, we can then set up the equations equal to each other and solve for the unknown, which is time.

Solution

1. Equation for the gazelle: The gazelle has a constant velocity, which is its average velocity, since it is not accelerating. Therefore, we use Equation 3.10 with $x_0=0$:

   $$\mathrm{ten}={\mathrm{ten}}_0+\stackrel{\text{–}}{v}t=\stackrel{\text{–}}{v}t.$$

   Equation for the cheetah: The chetah is accelerating from residue, so we use Equation three.13 with $x_0=0$ and ${\mathrm{five}}_0=0$:

   $$x=x_0+v_{0}t+\frac{\mathrm{one}}{\mathrm{two}}a{t}^{\mathrm{ii}}=\frac{\mathrm{i}}{2}a{t}^2.$$

   Now we have an equation of motion for each animal with a common parameter, which can be eliminated to detect the solution. In this example, nosotros solve for t:

   $$\begin{array}{}\\ \\ x=\stackrel{\text{–}}{v}t=\frac{1}{2}a{t}^2\hfill \\ t=\frac{\mathrm{ii}\stackrel{\text{–}}{\mathrm{five}}}{a}.\hfill \end{array}$$

   The gazelle has a constant velocity of 10 m/s, which is its average velocity. The acceleration of the chetah is 4 yard/s². Evaluating t, the fourth dimension for the cheetah to attain the gazelle, we have

   $$t=\frac{2\stackrel{\text{–}}{v}}{a}=\frac{2(\mathrm{x\; k/south})}{4{\text{yard/s}}^2}=5\text{s}\text{.}$$

2. To become the deportation, nosotros use either the equation of motion for the cheetah or the gazelle, since they should both give the same reply.
   Deportation of the cheetah:

   $$\mathrm{ten}=\frac{1}{2}a{t}^2=\frac{1}{2}(\mathrm{four}{\text{chiliad/south}}^2){(\mathrm{v})}^2=50\text{m}\text{.}$$

   Deportation of the gazelle:

   $$x=\stackrel{\text{–}}{v}t=\mathrm{10\; m/s}(5)=50\text{yard}\text{.}$$

   We see that both displacements are equal, every bit expected.

Significance

Information technology is of import to clarify the motion of each object and to use the appropriate kinematic equations to describe the individual motion. It is also important to have a expert visual perspective of the two-trunk pursuit problem to see the common parameter that links the movement of both objects.

Check Your Understanding 3.half dozen

A bicycle has a abiding velocity of ten 1000/s. A person starts from remainder and begins to run to catch up to the bicycle in 30 s when the bicycle is at the same position every bit the person. What is the acceleration of the person?

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Source: https://openstax.org/books/university-physics-volume-1/pages/3-4-motion-with-constant-acceleration

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