How To Find Initial Velocity With Time And Acceleration

Learning Objectives

By the stop of this section, you will be able to:

• Identify which equations of motility are to exist used to solve for unknowns.
• Use advisable equations of motion to solve a two-body pursuit problem.

You might estimate that the greater the acceleration of, say, a automobile moving away from a cease sign, the greater the car's displacement in a given time. But, we have not developed a specific equation that relates acceleration and deportation. In this section, we look at some user-friendly 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 ii objects, called two-body pursuit problems.

Notation

First, allow usa brand some simplifications in annotation. Taking the initial fourth dimension to be zero, as if time is measured with a stopwatch, is a dandy simplification. Since elapsed fourth dimension is $\text{Δ}t=t_{\text{f}}-t_0$ , taking $t_0=0$ means that $\text{Δ}t=t_{\text{f}}$ , the last time on the stopwatch. When initial time is taken to be aught, nosotros use 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 final position, and v is the concluding velocity. This gives a simpler expression for elapsed fourth dimension, $\text{Δ}t=t$ . It also simplifies the expression for x displacement, which is at present $\text{Δ}x=x-x_0$ . Also, it simplifies the expression for change in velocity, which is now $\text{Δ}\mathrm{five}=v-v_0$ . To summarize, using the simplified annotation, with the initial time taken to be zero,

$$\begin{array}{c}\text{Δ}t=t\hfill \\ \text{Δ}x=x-{\mathrm{10}}_0\hfill \\ \text{Δ}v=\mathrm{five}-v_0,\hfill \end{array}$$

where the subscript 0 denotes an initial value and the absence of a subscript denotes a concluding value in any motility is under consideration.

We now 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,

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

Thus, we tin use the symbol a for acceleration at all times. Bold dispatch to exist constant does not seriously limit the situations we can study nor does information technology degrade the accuracy of our handling. For one thing, dispatch is constant in a keen number of situations. Furthermore, in many other situations nosotros can describe motion accurately past assuming a abiding acceleration equal to the average dispatch for that motion. Lastly, for motion during which acceleration changes drastically, such equally a car accelerating to superlative speed and and so braking to a stop, motion tin can exist considered in separate parts, each of which has its own constant acceleration.

Displacement and Position from Velocity

To get our first ii equations, nosotros start with the definition of average velocity:

$$\stackrel{\text{–}}{5}=\frac{\text{Δ}x}{\text{Δ}t}.$$

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

$$\stackrel{\text{–}}{v}=\frac{x-{\mathrm{10}}_0}{t}.$$

Solving for x gives united states of america

where the average velocity is

The equation $\stackrel{\text{–}}{v}=\frac{v_0+v}{2}$ reflects the fact that when dispatch is constant, $\stackrel{\text{–}}{v}$ is simply the elementary average of the initial and final velocities. Figure iii.18 illustrates this concept graphically. In part (a) of the effigy, acceleration is constant, with velocity increasing at a constant rate. The average velocity during the ane-h interval from twoscore km/h to 80 km/h is lx km/h:

$$\stackrel{\text{–}}{5}=\frac{{\mathrm{five}}_0+v}{\mathrm{two}}=\frac{\mathrm{xl}\text{km/h}+\mathrm{fourscore}\text{km/h}}{\mathrm{two}}=60\text{km/h}\text{.}$$

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

Figure 3.18 (a) Velocity-versus-time graph with abiding acceleration showing the initial and final velocities $v_0\text{and}v$. The average velocity is $\frac{1}{\mathrm{ii}}(v_0+v)=60\text{km}\text{/}\text{h}$. (b) Velocity-versus-time graph with an acceleration that changes with time. The boilerplate velocity is not given past $\frac{1}{2}(v_0+\mathrm{five})$, just is greater than 60 km/h.

Solving for Terminal Velocity from Dispatch 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 annotation for $\text{Δ}v$ and $\text{Δ}t$ gives united states

$$a=\frac{v-5_0}{t}\left(\text{constant}a\right).$$

Solving for v yields

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

three.12

Example 3.seven

Calculating Concluding Velocity

An airplane lands with an initial velocity of 70.0 one thousand/due south and so accelerates opposite to the motion at 1.fifty yard/s² for 40.0 s. What is its final velocity?

Strategy

Outset, we identify the knowns: $v_0=\mathrm{seventy}\text{m/southward,}a=\mathrm{-1.fifty}{\text{m/s}}^2,t=40\text{south}$ .

Second, nosotros identify the unknown; in this instance, it is final velocity $v_{\text{f}}$ .

Last, we determine which equation to use. To do this we figure out which kinematic equation gives the unknown in terms of the knowns. Nosotros calculate the last velocity using Equation 3.12, $v=5_0+at$ .

Solution

Substitute the known values and solve:

$$v=v_0+at=70.0\text{m/s}+\left(\mathrm{-ane.50}\text{thousand/}{\text{southward}}^2\right)\left(\mathrm{forty.0\; south}\right)=\mathrm{10.0\; m/s.}$$

Figure 3.19 is a sketch that shows the acceleration and velocity vectors.

Effigy 3.nineteen The airplane lands with an initial velocity of 70.0 m/s and slows to a last velocity of ten.0 m/southward earlier heading for the terminal. Note the acceleration is negative because its direction is reverse to its velocity, which is positive.

Significance

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

In addition to being useful in problem solving, the equation $v=v_0+at$ gives us insight into the relationships amid velocity, acceleration, and time. We can run across, for case, that

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

All these observations fit our intuition. Note that it is always useful to examine basic equations in calorie-free of our intuition and feel to check that they practice indeed describe nature accurately.

Solving for Final Position with Constant Acceleration

We tin can combine the previous equations to find a tertiary equation that allows us to calculate the terminal position of an object experiencing constant dispatch. We start with

$$v=v_0+at.$$

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

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

Since $\frac{v_0+v}{\mathrm{ii}}=\stackrel{\text{–}}{5}$ for abiding acceleration, we have

$$\stackrel{\text{–}}{\mathrm{five}}=v_0+\frac{1}{\mathrm{ii}}at.$$

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

$$x={\mathrm{10}}_0+v_{0}t+\frac{\mathrm{i}}{\mathrm{two}}a{t}^2\left(\text{constant}a\right).$$

three.13

Example 3.eight

Calculating Displacement of an Accelerating Object

Dragsters can attain an average acceleration of 26.0 m/s². Suppose a dragster accelerates from remainder at this rate for v.56 s Figure iii.xx. How far does it travel in this time?

Figure 3.20 U.South. Regular army Summit Fuel pilot Tony "The Sarge" Schumacher begins a race with a controlled burnout. (credit: Lt. Col. William Thurmond. Photograph Courtesy of U.S. Army.)

Strategy

Start, let'south draw a sketch Figure 3.21. We are asked to find deportation, which is x if nosotros take $x_0$ to be zero. (Think nearly ${\mathrm{ten}}_0$ every bit the starting line of a race. It can be anywhere, but we call it nothing and measure out all other positions relative to information technology.) We can use the equation $\mathrm{10}=x_0+v_{0}t+\frac{\mathrm{i}}{\mathrm{ii}}a{t}^{\mathrm{ii}}$ when we identify $5_0$ , $a$ , and t from the statement of the problem.

Figure 3.21 Sketch of an accelerating dragster.

Solution

First, we need to identify the knowns. Starting from rest means that ${\mathrm{five}}_0=0$ , a is given as 26.0 m/s² and t is given equally 5.56 due south.

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

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

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

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

Substituting the identified values of a and t gives

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

Significance

If nosotros convert 402 yard to miles, we find that the distance covered is very close to one-quarter of a mile, the standard distance for elevate racing. So, our respond is reasonable. This is an impressive deportation to embrace in merely 5.56 southward, but superlative-notch dragsters tin can do a quarter mile in even less time than this. If the dragster were given an initial velocity, this would add another term to the distance equation. If the same acceleration and time are used in the equation, the altitude covered would be much greater.

What else tin we learn by examining the equation $\mathrm{10}=x_0+5_{0}t+\frac{\mathrm{i}}{2}a{t}^2?$ Nosotros can see the following relationships:

• Displacement depends on the foursquare of the elapsed time when acceleration is non zero. In Instance 3.8, the dragster covers only one-quaternary of the total distance in the first half of the elapsed time.
• If dispatch is zero, then initial velocity equals average velocity $({\mathrm{five}}_0=\stackrel{\text{–}}{v})$, and $x=x_0+v_{0}t+\frac{\mathrm{one}}{\mathrm{two}}a{t}^2\text{becomes}\mathrm{10}=x_0+{\mathrm{five}}_{0}t.$

Solving for Last Velocity from Distance and Acceleration

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

$$t=\frac{v-5_0}{a}.$$

Substituting this and $\stackrel{\text{–}}{5}=\frac{v_0+v}{2}$ into $x=x_0+\stackrel{\text{–}}{\mathrm{five}}t$ , we get

$$5^2=v_0^2+2a(\mathrm{ten}-x_0)(\text{constant}a).$$

3.14

Example three.9

Computing Final Velocity

Calculate the final velocity of the dragster in Example iii.8 without using data about time.

Strategy

The equation $v^2=v_0^{\mathrm{ii}}+2a(x-{\mathrm{ten}}_0)$ is ideally suited to this task because information technology relates velocities, acceleration, and displacement, and no time information is required.

Solution

Commencement, we identify the known values. We know that v ₀ = 0, since the dragster starts from balance. Nosotros also know that 10 − ten ₀ = 402 m (this was the answer in Example 3.8). The average acceleration was given past a = 26.0 m/s².

Second, we substitute the knowns into the equation $v^2={\mathrm{five}}_0^{\mathrm{ii}}+2a(\mathrm{ten}-x_0)$ and solve for v:

$$5^2=0+\mathrm{two}\left(26.0{\text{one thousand/s}}^{\mathrm{ii}}\right)\left(402\text{m}\right).$$

Thus,

$$\begin{array}{ccc}{v}^{\mathrm{ii}}\hfill & =\hfill & 2.09\times {10}^4{\text{one thousand}}^2{\text{/due south}}^2\hfill \\ \\ 5\hfill & =\hfill & \sqrt{\mathrm{ii.09}\times {10}^{\mathrm{iv}}{\text{grand}}^{\mathrm{ii}}{\text{/due south}}^{\mathrm{ii}}}=145\text{grand/s}\text{.}\hfill \end{array}$$

Significance

A velocity of 145 chiliad/s is about 522 km/h, or about 324 mi/h, only 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 point a velocity in the same direction every bit the acceleration.

An exam of the equation $v^2=5_0^2+2a(x-x_0)$ can produce additional 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 acceleration, a motorcar that is going twice as fast doesn't merely stop in twice the distance. It takes much farther to terminate. (This is why we have reduced speed zones nearly schools.)

Putting Equations Together

In the post-obit examples, we go on to explore ane-dimensional motility, but in situations requiring slightly more than algebraic manipulation. The examples also give insight into problem-solving techniques. The note that follows is provided for piece of cake reference to the equations needed. Be aware that these equations are not contained. In many situations we have two unknowns and need two equations from the fix to solve for the unknowns. Nosotros need as many equations every bit in that location are unknowns to solve a given situation.

Summary of Kinematic Equations (constant a)

$$x=x_0+\stackrel{\text{–}}{v}t$$

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

$$v=v_0+at$$

$$\mathrm{10}={\mathrm{ten}}_0+{\mathrm{five}}_{0}t+\frac{1}{2}a{t}^2$$

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

Before we go into the examples, allow'due south await at some of the equations more closely to see the behavior of acceleration at extreme values. Rearranging Equation 3.12, we accept

$$a=\frac{5-v_0}{t}.$$

From this we meet that, for a finite time, if the departure between the initial and final velocities is small, the acceleration is pocket-sized, budgeted zilch in the limit that the initial and final velocities are equal. On the contrary, in the limit $t\to 0$ for a finite deviation between the initial and last velocities, dispatch becomes space.

Similarly, rearranging Equation three.14, we can express acceleration in terms of velocities and displacement:

$$a=\frac{v^2-v_0^2}{2(x-{\mathrm{ten}}_0)}.$$

Thus, for a finite difference between 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 concluding velocities approaches zippo for a finite displacement.

Example iii.ten

How Far Does a Automobile Go?

On dry concrete, a car can accelerate opposite to the motion at a rate of vii.00 m/s², whereas on moisture concrete information technology tin accelerate opposite to the motion at simply v.00 m/south². Observe the distances necessary to stop a car moving at 30.0 chiliad/s (about 110 km/h) on (a) dry concrete and (b) moisture concrete. (c) Repeat both calculations and find the deportation from the point where the commuter sees a traffic calorie-free plow red, taking into business relationship his reaction time of 0.500 southward to get his foot on the brake.

Strategy

First, we demand to draw a sketch Effigy 3.22. To decide which equations are best to utilise, we demand to list all the known values and identify exactly what we demand to solve for.

Figure 3.22 Sample sketch to visualize acceleration opposite to the motion and stopping altitude of a car.

Solution

1. Commencement, nosotros need to place the knowns and what nosotros desire to solve for. We know that v ₀ = thirty.0 g/south, v = 0, and a = −7.00 m/s² (a is negative because it is in a direction opposite to velocity). We take 10 ₀ to be zero. We are looking for displacement $\text{Δ}x$, or 10 − ten ₀.
   2d, we place the equation that will assist us solve the problem. The best equation to use is

   $$5^{\mathrm{ii}}=v_0^{\mathrm{two}}+\mathrm{two}a(x-x_0).$$

   This equation is best considering it includes simply one unknown, ten. We know the values of all the other variables in this equation. (Other equations would allow us to solve for x, but they require us to know the stopping fourth dimension, t, which we practise non know. Nosotros could use them, but it would entail additional calculations.)
   Third, we rearrange the equation to solve for x:

   $$\mathrm{10}-{\mathrm{ten}}_0=\frac{{\mathrm{five}}^2-v_0^{\mathrm{two}}}{2a}$$

   and substitute the known values:

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

   Thus,

   $$\mathrm{10}=\mathrm{64.three}\text{m on dry physical}\text{.}$$

2. This part tin exist solved in exactly the aforementioned manner as (a). The only deviation is that the dispatch is −5.00 one thousand/due south². The outcome is

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

3. When the driver reacts, the stopping distance is the same equally it is in (a) and (b) for dry out and wet concrete. Then, to answer this question, nosotros need to calculate how far the car travels during the reaction time, and then add together that to the stopping fourth dimension. It is reasonable to presume the velocity remains constant during the driver'southward reaction fourth dimension.
   To do this, we, again, identify the knowns and what nosotros desire to solve for. Nosotros know that $\stackrel{\text{–}}{v}=30.0\text{m/s}$, $t_{\text{reaction}}=0.500\text{southward}$, and $a_{\text{reaction}}=0$. We have ${\mathrm{10}}_{\text{0-reaction}}$ to be zero. Nosotros are looking for ${\mathrm{10}}_{\text{reaction}}$.
   Second, as before, we place the best equation to employ. In this case, $\mathrm{10}={\mathrm{ten}}_0+\stackrel{\text{–}}{5}t$ works well because the simply unknown value is x, which is what we want to solve for.
   3rd, we substitute the knowns to solve the equation:

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

   This ways the automobile travels 15.0 m while the driver reacts, making the total displacements in the 2 cases of dry and wet concrete 15.0 yard greater than if he reacted instantly.
   Concluding, we then add the deportation during the reaction fourth dimension to the displacement when braking (Figure 3.23),

   $${\mathrm{10}}_{\text{braking}}+{\mathrm{ten}}_{\text{reaction}}=x_{\text{total}},$$

   and observe (a) to be 64.3 thousand + 15.0 m = 79.three m when dry and (b) to be ninety.0 yard + fifteen.0 1000 = 105 m when wet.

Figure 3.23 The distance necessary to stop a car varies greatly, depending on road conditions and driver reaction time. Shown here are the braking distances for dry and wet pavement, as calculated in this case, for a machine traveling initially at xxx.0 thou/s. Besides shown are the total distances traveled from the bespeak when the driver offset sees a light turn red, assuming a 0.500-due south reaction fourth dimension.

Significance

The displacements establish in this instance seem reasonable for stopping a fast-moving machine. It should have longer to stop a machine on wet pavement than dry out. Information technology is interesting that reaction time adds significantly to the displacements, only more important is the general approach to solving problems. We identify the knowns and the quantities to be determined, then find an appropriate equation. If there is more than than one unknown, nosotros need as many independent equations as in that location are unknowns to solve. At that place is oft more than one way to solve a problem. The various parts of this case can, in fact, be solved past other methods, but the solutions presented here are the shortest.

Case 3.eleven

Calculating Time

Suppose a car merges into freeway traffic on a 200-m-long ramp. If its initial velocity is ten.0 m/s and it accelerates at 2.00 1000/s², how long does information technology take the automobile to travel the 200 one thousand upwardly the ramp? (Such information might be useful to a traffic engineer.)

Strategy

Start, we depict a sketch Effigy three.24. We are asked to solve for time t. As before, we identify the known quantities to choose a user-friendly physical relationship (that is, an equation with one unknown, t.)

Figure 3.24 Sketch of a car accelerating on a freeway ramp.

Solution

Once more, we identify the knowns and what we want to solve for. We know that $x_0=0,$
$v_0=10\text{m/s},a=2.00\text{k/}{\text{due south}}^{\mathrm{ii}}$ , and x = 200 m.

Nosotros need to solve for t. The equation $\mathrm{10}=x_0+v_{0}t+\frac{\mathrm{i}}{2}a{t}^2$ works best because the only unknown in the equation is the variable t, for which we need to solve. From this insight we see that when we input the knowns into the equation, we stop up with a quadratic equation.

We need to rearrange the equation to solve for t, then substituting the knowns into the equation:

$$200\text{g}=0\text{m}+(\mathrm{ten.0}\text{g/southward})t+\frac{\mathrm{ane}}{\mathrm{two}}\left(2.00{\text{g/s}}^{\mathrm{ii}}\right)t^2.$$

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

$$200=10t+t^{\mathrm{two}}.$$

We then use the quadratic formula to solve for t,

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

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

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

Significance

Whenever an equation contains an unknown squared, there are two solutions. In some problems both solutions are meaningful; in others, only i solution is reasonable. The ten.0-s answer seems reasonable for a typical freeway on-ramp.

Cheque Your Understanding 3.5

A rocket accelerates at a rate of twenty k/southward² during launch. How long does it take the rocket to achieve a velocity of 400 m/s?

Example 3.12

Dispatch of a Spaceship

A spaceship has left Earth'southward orbit and is on its way to the Moon. It accelerates at 20 m/s² for 2 min and covers a altitude of g km. What are the initial and final velocities of the spaceship?

Strategy

We are asked to find the initial and last velocities of the spaceship. Looking at the kinematic equations, we meet that i equation will not requite the respond. Nosotros must utilise i kinematic equation to solve for one of the velocities and substitute it into another kinematic equation to become the second velocity. Thus, we solve two of the kinematic equations simultaneously.

Solution

Outset nosotros solve for $v_0$ using $x=x_0+v_{0}t+\frac{1}{\mathrm{ii}}a{t}^2:$

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

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

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

And so nosotros substitute $v_0$ into $5={\mathrm{five}}_0+at$ to solve for the final velocity:

$$v=5_0+at=7133.3\text{m/s}+(20.0{\text{m/s}}^2)(120.0\text{s})=9533.3\text{m/due south.}$$

Significance

There are six variables in displacement, time, velocity, and acceleration that describe movement in ane dimension. The initial conditions of a given trouble can be many combinations of these variables. Because of this diversity, solutions may not be every bit easy every bit simple substitutions into ane of the equations. This example illustrates that solutions to kinematics may require solving 2 simultaneous kinematic equations.

With the basics of kinematics established, we can go on to many other interesting examples and applications. In the process of developing kinematics, nosotros have as well glimpsed a general approach to trouble solving that produces both right answers and insights into physical relationships. The side by side level of complication in our kinematics problems involves the motion of two interrelated bodies, called two-body pursuit bug.

Two-Trunk Pursuit Problems

Up until this indicate nosotros have looked at examples of motion involving a unmarried body. Fifty-fifty for the problem with two cars and the stopping distances on moisture and dry roads, nosotros divided this trouble into 2 separate problems to find the answers. In a ii-body pursuit problem, the motions of the objects are coupled—pregnant, the unknown we seek depends on the motion of both objects. To solve these problems nosotros write the equations of motion for each object and then solve them simultaneously to notice the unknown. This is illustrated in Figure 3.25.

Figure 3.25 A two-body pursuit scenario where car 2 has a constant velocity and car 1 is behind with a abiding dispatch. Car one catches up with car 2 at a later on time.

The fourth dimension and distance required for machine 1 to catch car 2 depends on the initial altitude car 1 is from car 2 equally well as the velocities of both cars and the acceleration of car 1. The kinematic equations describing the motion of both cars must be solved to observe these unknowns.

Consider the post-obit instance.

Example 3.13

Cheetah Communicable a Gazelle

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

Strategy

We use the prepare of equations for constant dispatch to solve this problem. Since there are two objects in motion, we accept carve up equations of motion describing each animate being. But what links the equations is a common parameter that has the same value for each animal. If nosotros wait at the problem closely, it is articulate the mutual parameter to each fauna is their position 10 at a later time t. Since they both showtime at $x_0=0$ , their displacements are the same at a afterward fourth dimension t, when the chetah catches up with the gazelle. If nosotros pick the equation of motion that solves for the displacement for each animate being, we can then prepare 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 boilerplate velocity, since it is non accelerating. Therefore, we employ Equation 3.x with $x_0=0$:

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

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

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

   Now we accept an equation of motion for each animal with a common parameter, which can be eliminated to notice the solution. In this case, we solve for t:

   $$\begin{array}{}\\ \\ \mathrm{10}=\stackrel{\text{–}}{5}t=\frac{1}{2}a{t}^2\hfill \\ t=\frac{2\stackrel{\text{–}}{v}}{a}.\hfill \end{array}$$

   The gazelle has a constant velocity of 10 g/s, which is its boilerplate velocity. The acceleration of the cheetah is 4 m/due south². Evaluating t, the time for the cheetah to accomplish the gazelle, we have

   $$t=\frac{\mathrm{ii}\stackrel{\text{–}}{v}}{a}=\frac{2(\mathrm{ten\; m/s})}{\mathrm{iv}{\text{one thousand/due south}}^{\mathrm{ii}}}=5\text{south}\text{.}$$

2. To become the displacement, we utilise either the equation of movement for the chetah or the gazelle, since they should both give the same answer.
   Displacement of the chetah:

   $$\mathrm{10}=\frac{1}{2}a{t}^{\mathrm{ii}}=\frac{\mathrm{ane}}{\mathrm{ii}}(4{\text{grand/s}}^2){(5)}^2=50\text{m}\text{.}$$

   Displacement of the gazelle:

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

   We run across that both displacements are equal, as expected.

Significance

It is important to analyze the motion of each object and to use the appropriate kinematic equations to describe the individual move. Information technology is also important to accept a skilful visual perspective of the two-torso pursuit problem to see the common parameter that links the movement of both objects.

Check Your Agreement 3.vi

A bicycle has a constant velocity of 10 k/s. A person starts from residual and begins to run to catch up to the bike in 30 s when the bicycle is at the aforementioned position as the person. What is the acceleration of the person?

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

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