Can an Object With Constant Acceleration Reverse Its Direction

Scholarship Objectives

Past the end of this section, you will be able to:

• Identify which equations of motion are to be victimized to solve for unknowns.
• Employ allow equations of motion to solve a two-body pursual job.

You might venture that the greater the acceleration of, say, a car moving away from a stop sign, the greater the machine's displacement in a tending time. But, we have non developed a specialised equation that relates acceleration and displacement. In this department, we look at some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration. We first investigate a single object in motion, called single-body motion. Then we investigate the motion of two objects, called two-body quest problems.

Notation

First, let us prepar some simplifications in notation. Attractive the first time to be zero, as if time is measured with a stopo watch, is a expectant reduction. Since elapsed time is $\text{Δ}t=t_{\text{f}}-t_0$ , taking $t_0=0$ means that $\text{Δ}t=t_{\text{f}}$ , the final prison term on the stopwatch. When first time is taken to be zero, we use the subscript 0 to announce initial values of position and velocity. That is, $x_0$ is the initial put back and $v_0$ is the first velocity. We put no subscripts on the final values. That is, t is the final examination time, x is the final position, and v is the unalterable velocity. This gives a simpler verbal expression for elapsed time, $\text{Δ}t=t$ . It also simplifies the expression for x displacement, which is now $\text{Δ}x=x-x_0$ . Also, it simplifies the expression for alter in velocity, which is now $\text{Δ}v=v-v_0$ . To summarise, using the simplified notation, with the first time taken to be zero,

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

where the subscript 0 denotes an initial valuate and the absence of a subscript denotes a final value in whatever question is relevant.

We now make the grave assumption that acceleration is constant. This 15-Aug allows us to avoid using calculus to ascertain instant acceleration. Since acceleration is constant, the average and instantaneous accelerations are equal—that is,

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

Thus, we can use the symbolisation a for acceleration at all times. Assuming acceleration to be constant quantity does non badly circumscribe the situations we can study nor does it demean the accuracy of our treatment. For one thing, acceleration is constant in a great number of situations. Furthermore, in many other situations we can discover motion accurately by assuming a constant acceleration close to the average acceleration for that motion. Last, for motion during which speedup changes drastically, such as a car accelerating to overstep hurrying then braking to a hitch, motion can be advised in separate parts, apiece of which has its own uninterrupted acceleration.

Displacement and Position from Velocity

To perplex our initiatory two equations, we start with the definition of average velocity:

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

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

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

Solving for x gives us

where the average velocity is

The equation $\stackrel{\text{–}}{v}=\frac{v_0+v}{2}$ reflects the fact that when acceleration is constant, $\stackrel{\text{–}}{v}$ is just the simple average of the initial and final velocities. Figure 3.18 illustrates this concept graphically. In break u (a) of the figure, acceleration is invariant, with velocity progressive at a constant grade. The average velocity during the 1-h interval from 40 kilometer/h to 80 klick/h is 60 kilometres per hour:

$$\stackrel{\text{–}}{v}=\frac{v_0+v}{2}=\frac{40\text{kilometers per hour}+80\text{km/h}}{2}=60\text{km/h}\text{.}$$

In split up (b), acceleration is not constant. During the 1-h interval, velocity is finisher to 80 km/h than 40 km/h. Thus, the middling speed is greater than in part (a).

Figure 3.18 (a) Velocity-versus-time graph with constant quickening screening the initial and final velocities $v_0\text{and}v$. The mediocre velocity is $\frac{1}{2}(v_0+v)=60\text{km}\text{/}\text{h}$. (b) Velocity-versus-time graph with an acceleration that changes with time. The average speed is not given away $\frac{1}{2}(v_0+v)$, but is greater than 60 klick/h.

Solving for Final Velocity from Acceleration and Time

We can derive another useful equivalence by manipulating the definition of acceleration:

$$a=\frac{\text{Δ}v}{\text{Δ}t}.$$

Substituting the simplified notational system for $\text{Δ}v$ and $\text{Δ}t$ gives us

$$a=\frac{v-v_0}{t}\left(\text{constant}a\right).$$

Solving for v yields

$$v=v_0+at\left(\text{perpetual}a\right).$$

3.12

Example 3.7

Hard Final Speed

An plane lands with an initial speed of 70.0 m/s and then accelerates opposite to the move at 1.50 m/s² for 40.0 s. What is its inalterable velocity?

Strategy

First, we identify the knowns: $v_0=70\text{m/s,}a=\mathrm{-1.50}{\text{m/s}}^2,t=40\text{s}$ .

Second, we key the unknown; in that case, it is final speed $v_{\text{f}}$ .

Last, we determine which equation to use. To do this we figure impermissible which kinematic equation gives the unknown in terms of the knowns. We calculate the final speed using Equation 3.12, $v=v_0+at$ .

Result

Replacement the known values and solve:

$$v=v_0+at=70.0\text{m/s}+\left(\mathrm{-1.50}\text{m/}{\text{s}}^2\right)\left(\mathrm{40.0\; s}\right)=\mathrm{10.0\; m/s.}$$

Figure 3.19 is a resume that shows the acceleration and speed vectors.

Fig 3.19 The airplane lands with an initial velocity of 70.0 m/s and slows to a final velocity of 10.0 m/s ahead heading for the closing. Short letter the acceleration is antagonistic because its way is inverse to its speed, which is positive.

Significance

The final velocity is much less than the initial velocity, as in demand when slowing down, only is still certain (go out figure). With jet engines, reverse thrust can be kept up long plenty to stop the plane and lead off moving IT backward, which is indicated by a dissentient final speed, but is not the case here.

In increase to being useful in problem solving, the equation $v=v_0+at$ gives us insight into the relationships among velocity, quickening, and time. We give the axe see, for example, that

• Final velocity depends on how large the quickening is you said it longish it lasts
• If the acceleration is zero, then the unalterable velocity equals the first velocity (v = v ₀), As expected (put differently, speed is unfailing)
• If a is negative, and then the final velocity is less than the initial velocity

All these observations fit our intuition. Note that IT is ever utilitarian to examine basic equations in twinkle of our hunch and experience to tab that they DO indeed describe nature accurately.

Solving for Final Position with Constant Speedup

We fundament combine the previous equations to find a one-third equation that allows us to calculate the final position of an object experiencing unflagging acceleration. We start with

$$v=v_0+at.$$

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

$$\frac{v_0+v}{2}=v_0+\frac{1}{2}at.$$

Since $\frac{v_0+v}{2}=\stackrel{\text{–}}{v}$ for unvarying acceleration, we have

$$\stackrel{\text{–}}{v}=v_0+\frac{1}{2}at.$$

Now we substitute this verbalism for $\stackrel{\text{–}}{v}$ into the equation for displacement, $x=x_0+\stackrel{\text{–}}{v}t$ , yielding

$$x=x_0+v_{0}t+\frac{1}{2}a{t}^2\left(\text{unvarying}a\right).$$

3.13

Example 3.8

Calculating Displacement of an Fast Physical object

Dragsters can achieve an average acceleration of 26.0 m/s². Suppose a dragster accelerates from rest at this rate for 5.56 s Figure 3.20. How ALIR does it travel in this time?

Pattern 3.20 U.S. Army Top Fuel fly Tony "The Sarge" Schumacher begins a race with a limited burnout. (credit: Lt. Col. William Thurmond. Photo Courtesy of U.S. Army.)

Strategy

First, let's draw a adumbrate Work out 3.21. We are asked to ascertain displacement, which is x if we ask $x_0$ to be zero. (Toy with $x_0$ as the start of a race. It hind end be anyplace, but we call it zero and measure all other positions relation to it.) We can use the equating $x=x_0+v_{0}t+\frac{1}{2}a{t}^2$ when we identify $v_0$ , $a$ , and t from the affirmation of the trouble.

Envision 3.21 Sketch of an accelerating dragster.

Solvent

First, we call for to identify the knowns. Start from rest substance that $v_0=0$ , a is given as 26.0 m/s² and t is relinquished as 5.56 s.

Second, we substitute the known values into the equation to wor for the obscure:

$$x=x_0+v_{0}t+\frac{1}{2}a{t}^2.$$

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

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

Substituting the known values of a and t gives

$$x=\frac{1}{2}(26.0{\text{m/s}}^2){(5.56\text{s})}^2=402\text{m}\text{.}$$

Significance

If we convert 402 m to miles, we find that the distance splashed is very close to one-quarter of a mile, the standard distance for scuff racing. So, our answer is reasonable. This is an impressive displacement to breed in only 5.56 s, but top-mountain pass dragsters can do a 440 yards in even out fewer time than this. If the dragster were given an initial velocity, this would lend another term to the distance equivalence. If the same acceleration and metre are used in the equation, the distance covered would be much greater.

What else hindquarters we learn by examining the equivalence $x=x_0+v_{0}t+\frac{1}{2}a{t}^2?$ We john see the following relationships:

• Supplanting depends on the squared of the elapsed time when acceleration is not zero. In Example 3.8, the dragster covers exclusive one-fourth of the total distance in the foremost one-half of the elapsed clock.
• If acceleration is zero, then initial speed equals average velocity $(v_0=\stackrel{\text{–}}{v})$, and $x=x_0+v_{0}t+\frac{1}{2}a{t}^2\text{becomes}x=x_0+v_{0}t.$

Resolution for Final Velocity from Distance and Acceleration

A one-fourth useful equivalence can be obtained from another algebraic use of previous equations. If we solve $v=v_0+at$ for t, we come

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

Substituting this and $\stackrel{\text{–}}{v}=\frac{v_0+v}{2}$ into $x=x_0+\stackrel{\text{–}}{v}t$ , we get

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

3.14

Example 3.9

Calculating Final Velocity

Calculate the last velocity of the dragster in Example 3.8 without using info about time.

Strategy

The equality $v^2=v_0^2+2a(x-x_0)$ is ideally suited to this tax because it relates velocities, quickening, and displacement, and no time information is required.

Solution

First, we identify the known values. We know that v ₀ = 0, since the dragster starts from stay. We as wel know that x − x ₀ = 402 m (this was the result in Example 3.8). The average acceleration was given by a = 26.0 m/s².

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

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

Thus,

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

Significance

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

An test of the equality $v^2=v_0^2+2a(x-x_0)$ can produce additional insights into the general relationships among physical quantities:

• The final velocity depends connected how large the acceleration is and the distance finished which IT acts.
• For a fixed acceleration, a car that is going twice as fast doesn't just stop in twice the distance. IT takes more farther to stop. (This is why we have reduced speed zones near schools.)

Putting Equations Together

In the following examples, we persist in to explore combined-magnitude question, simply in situations requiring slightly more algebraic manipulation. The examples also give insight into problem-solving techniques. The mention that follows is provided for easy denotation to the equations needed. Be aware that these equations are non independent. In many situations we have cardinal unknowns and pauperism ii equations from the set to solve for the unknowns. We need as galore equations as there are unknowns to solve a given post.

Summary of Kinematic Equations (constant a)

$$x=x_0+\stackrel{\text{–}}{v}t$$

$$\stackrel{\text{–}}{v}=\frac{v_0+v}{2}$$

$$v=v_0+at$$

$$x=x_0+v_{0}t+\frac{1}{2}a{t}^2$$

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

Before we tangle with the examples, let's look at some of the equations more closely to see the demeanor of acceleration at extreme values. Rearranging Equivalence 3.12, we have

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

From this we see that, for a finite time, if the divergence between the initial and final velocities is small, the acceleration is small, approaching zero in the limit that the initial and final velocities are equal. To the contrary, in the set $t\to 0$ for a finite difference betwixt the initial and final velocities, acceleration becomes infinite.

Likewise, rearranging Equating 3.14, we can utter acceleration in terms of velocities and displacement:

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

So, for a finite difference between the initial and final velocities quickening becomes infinite in the limit the displacement approaches nada. Acceleration approaches zero in the limit the difference of opinion in initial and final velocities approaches zero for a finite displacement.

Example 3.10

How Out-of-the-way Does a Car Go?

On dry concrete, a car can quicken opposite to the motion at a rate of 7.00 m/s², whereas on wet concrete it fire accelerate opposite to the move at single 5.00 m/s². Find the distances obligatory to stop a car riding at 30.0 m/s (about 110 kilometres per hour) along (a) brut concrete and (b) bedewed tangible. (c) Repeat both calculations and find the supplanting from the point where the driver sees a traffic lightheaded turn red, fetching into account his reaction time of 0.500 s to baffle his foot on the brake.

Strategy

Opening, we need to draw a sketch Picture 3.22. To determine which equations are best to habituate, we need to leaning all the known values and identify exactly what we need to solve for.

Figure 3.22 Sample cartoon to visualize acceleration opposite to the motion and stopping distance of a car.

Solution

1. Opening, we need to name the knowns and what we want to solve for. We cognize that v ₀ = 30.0 m/s, v = 0, and a = −7.00 m/s² (a is negative because IT is in a direction opposite to velocity). We take x ₀ to be zero. We are looking for displacement $\text{Δ}x$, Oregon x − x ₀.
   Second, we discover the equation that will helper US solve the trouble. The best equality to habituate is

   $$v^2=v_0^2+2a(x-x_0).$$

   This equation is best because information technology includes solitary unity unknown, x. We be intimate the values of all the separate variables in this equation. (Unusual equations would allow us to solve for x, but they expect us to know the stopping prison term, t, which we do not do it. We could use them, but IT would entail additional calculations.)
   Third gear, we rearrange the equation to figure out for x:

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

   and substitute the known values:

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

   Thus,

   $$x=64.3\text{m on kiln-dried existent}\text{.}$$

2. This part can be solved in precisely the unchanged manner as (a). The lonesome difference is that the acceleration is −5.00 m/s². The result is

   $$x_{\text{sticky}}=90.0\text{m on wet factual.}$$

3. When the driver reacts, the stopping distance is the same arsenic it is in (a) and (b) for dry and wet concrete. So, to answer this inquiry, we need to calculate how far the car travels during the reaction time, and then add that to the stopping time. It is commonsense to assume the velocity corpse constant during the driver's response time.
   To do this, we, again, identify the knowns and what we require to solve for. We know that $\stackrel{\text{–}}{v}=30.0\text{m/s}$, $t_{\text{response}}=0.500\text{s}$, and $a_{\text{chemical reaction}}=0$. We take $x_{\text{0-chemical reaction}}$ to be nix. We are looking for $x_{\text{reaction}}$.
   Second, as before, we identify the best equation to utilise. In that case, $x=x_0+\stackrel{\text{–}}{v}t$ works well because the only unknown value is x, which is what we privation to solve for.
   Third, we substitute the knowns to solve the equality:

   $$x=0+\left(30.0\text{m/s}\right)\left(0.500\text{s}\right)=15.0\text{m}.$$

   This means the car travels 15.0 m while the driver reacts, making the total displacements in the 2 cases of dry and wet concrete 15.0 m greater than if he reacted outright.
   Last, we then add the displacement during the chemical reaction time to the supplanting when braking (See 3.23),

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

   and notic (a) to equal 64.3 m + 15.0 m = 79.3 m when dried and (b) to embody 90.0 m + 15.0 m = 105 m when wet.

Form 3.23 The distance necessary to blockade a car varies greatly, depending on route conditions and driver reaction time. Shown here are the braking distances for dry and wet pavement, every bit calculated in this example, for a car traveling initially at 30.0 m/s. Also shown are the total distances cosmopolitan from the channelis when the driver first sees a light turn red, assuming a 0.500-s reaction sentence.

Significance

The displacements found in this deterrent example seem reasonable for stopping a fast-moving car. It should learn thirster to closure a car on wet pavement than dry. It is interesting that response time adds significantly to the displacements, but more important is the general approach shot to solving problems. We identify the knowns and the quantities to be determined, then find an appropriate equating. If in that location is more than one strange, we need arsenic many strong-minded equations as there are unknowns to clear. There is often much one way to solve a problem. The various parts of this example behind, in fact, be solved by other methods, but the solutions presented here are the shortest.

Example 3.11

Calculating Time

Suppose a car merges into freeway traffic happening a 200-m-long storm. If its first speed is 10.0 m/s and it accelerates at 2.00 m/s², how long does IT take the railcar to traveling the 200 m up the ramp? (Such information might be useful to a traffic engineer.)

Strategy

First, we haul a survey Figure 3.24. We are asked to solve for sentence t. Atomic number 3 before, we identify the known quantities to choose a convenient physical relationship (that is, an equation with one unknown, t.)

Figure 3.24 Sketch of a car fast happening a pike ramp.

Solution

Again, we name the knowns and what we deficiency to work out for. We acknowledge that $x_0=0,$
$v_0=10\text{m/s},a=2.00\text{m/}{\text{s}}^2$ , and x = 200 m.

We need to solve for t. The equality $x=x_0+v_{0}t+\frac{1}{2}a{t}^2$ whole works best because the only unknown in the equality is the variable t, for which we need to solve. From this insight we ascertain that when we input the knowns into the equating, we wind up with a quadratic.

We need to rearrange the equivalence to wor for t, then subbing the knowns into the equation:

$$200\text{m}=0\text{m}+(10.0\text{m/s})t+\frac{1}{2}\left(2.00{\text{m/s}}^2\right)t^2.$$

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

$$200=10t+t^2.$$

We then use the quadratic formula to solve for t,

$$\begin{array}{c}{t}^2+10t-200=0\\ \\ \\ t=\frac{\text{−}b\pm \sqrt{b^2-4ac}}{2a},\end{array}$$

which yields two solutions: t = 10.0 and t = −20.0. A negative value for time is unreasonable, since it would mean the issue happened 20 s before the question began. We can discard that solution. Thus,

$$t=10.0\text{s}\text{.}$$

Implication

Whenever an equation contains an unknown squared, there are deuce solutions. In some problems both solutions are meaningful; in others, only one solution is reasonable. The 10.0-s answer seems just for a typical state highway on-Allium tricoccum.

Check Your Understanding 3.5

A rocket accelerates at a charge per unit of 20 m/s² during launch. How long does it take the Eruca sativa to hit a speed of 400 m/s?

Example 3.12

Acceleration of a Spaceship

A spaceship has left Earth's reach and is on its way to the Moon. It accelerates at 20 m/s² for 2 min and covers a length of 1000 km. What are the first and final velocities of the spaceship?

Scheme

We are asked to find the initial and final velocities of the starship. Looking at the kinematic equations, we attend that one equation will not give the reply. We essential use ane kinematic equation to solve for one of the velocities and relief it into another kinematic equation to get the second velocity. Thus, we solve two of the kinematic equations simultaneously.

Solution

Beginning we work out for $v_0$ using $x=x_0+v_{0}t+\frac{1}{2}a{t}^2:$

$$x-x_0=v_{0}t+\frac{1}{2}a{t}^2$$

$$1.0\times {10}^6\text{m}=v_0(120.0\text{s})+\frac{1}{2}(20.0{\text{m/s}}^2){(120.0\text{s})}^2$$

$$v_0=7133.3\text{m/s}\text{.}$$

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

$$v=v_0+at=7133.3\text{m/s}+(20.0{\text{m/s}}^2)(120.0\text{s})=9533.3\text{m/s.}$$

Significance

There are six variables in displacement, metre, speed, and acceleration that describe motion in one dimension. The initial conditions of a surrendered job can be many combinations of these variables. Because of this diversity, solutions may not be As easy as simple substitutions into ace of the equations. This object lesson illustrates that solutions to kinematics may require solving two co-occurrent kinematic equations.

With the basics of kinematics established, we can march on to many another interesting examples and applications. In the process of developing kinematics, we throw also glimpsed a general approach to trouble resolution that produces both correct answers and insights into physical relationships. The next degree of complexity in our kinematics problems involves the motion of two interrelated bodies, called ii-body pursuit problems.

Two-Body Pursuit Problems

Up until this point we have looked at examples of motion involving a single body. True for the problem with two cars and the stopping distances connected wet and dry roads, we divided this problem into two secernate problems to find the answers. In a two-body pursuit problem, the motions of the objects are coupled—significance, the obscure we seek depends along the motion of both objects. To solve these problems we write the equations of motion for each object and then puzzle out them at the same time to find the unknown. This is illustrated in Figure 3.25.

Figure 3.25 A two-body pursuit scenario where car 2 has a stable velocity and car 1 is behind with a unvarying acceleration. Car 1 catches up with car 2 at a after time.

The clock and aloofness needful for railroad car 1 to catch car 2 depends on the initial aloofness car 1 is from railway car 2 every bit well A the velocities of some cars and the quickening of car 1. The kinematic equations describing the motion of both cars moldiness exist solved to find these unknowns.

Consider the succeeding example.

Example 3.13

Chetah Contagious a Gazelle

A cheetah waits in hiding behind a bush. The cheetah spots a gazelle linear past at 10 m/s. At the insistent the gazelle passes the cheetah, the cheetah accelerates from rest at 4 m/s² to take in the gazelle. (a) How long does it take the cheetah to catch the gazelle? (b) What is the displacement of the gazelle and cheetah?

Scheme

We usage the set of equations for stable acceleration to wor this problem. Since there are two objects in motion, we have separate equations of motion describing each mammal-like. But what links the equations is a common parametric quantity that has the synoptical value for each animal. If we consider the problem closely, it is clear the park parameter to each animal is their position x at a later meter t. Since they both head start at $x_0=0$ , their displacements are the same at a later time t, when the cheetah catches up with the gazelle. If we pick up the equating of gesture that solves for the displacement for from each one animal, we can then ordered the equations capable each other and wor for the unacknowledged, which is time.

Solution

1. Par for the gazelle: The gazelle has a constant velocity, which is its average velocity, since it is not fast. Thus, we wont Equation 3.10 with $x_0=0$:

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

   Equation for the cheetah: The cheetah is accelerating from rest, so we habit Equation 3.13 with $x_0=0$ and $v_0=0$:

   $$x=x_0+v_{0}t+\frac{1}{2}a{t}^2=\frac{1}{2}a{t}^2.$$

   Now we have an equation of motion for each animal with a popular parametric quantity, which bottom be eliminated to find the solution. In this lawsuit, we solve for t:

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

   The gazelle has a constant velocity of 10 m/s, which is its moderate speed. The acceleration of the cheetah is 4 m/s². Evaluating t, the time for the cheetah to strive the gazelle, we make

   $$t=\frac{2\stackrel{\text{–}}{v}}{a}=\frac{2(\mathrm{10\; m/s})}{4{\text{m/s}}^2}=5\text{s}\text{.}$$

2. To get the displacement, we employment either the equation of motion for the Acinonyx jubatus Oregon the gazelle, since they should both pass on the same answer.
   Displacement of the cheetah:

   $$x=\frac{1}{2}a{t}^2=\frac{1}{2}(4{\text{m/s}}^2){(5)}^2=50\text{m}\text{.}$$

   Displacement of the gazelle:

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

   We attend that both displacements are tight, as expected.

Significance

It is important to analyze the motion of each object and to use the right kinematic equations to describe the individual motion. Information technology is also principal to have a nifty visual perspective of the two-body pursuit problem to see the common parametric quantity that links the motion of some objects.

Check Your Understanding 3.6

A bicycle has a staunch velocity of 10 m/s. A soul starts from rest and begins to run to catch up to the bicycle in 30 s when the bicycle is at the same position As the person. What is the acceleration of the person?

Can an Object With Constant Acceleration Reverse Its Direction

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

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