What Does the Scale Read (in N) When the Elevator Is Accelerating Upward at 2.3 M/s2 ?

Learning Objectives

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

• Place which equations of motion are to be used to solve for unknowns.
• Utilize appropriate equations of motion to solve a two-torso pursuit problem.

You might guess that the greater the acceleration of, say, a car moving away from a stop sign, the greater the car's displacement in a given fourth dimension. But, we have non developed a specific 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 dispatch. We first investigate a single object in motion, called single-body motion. Then we investigate the motion of two objects, called two-body pursuit problems.

Notation

Outset, let the states make some simplifications in notation. Taking the initial time to be zero, as if time is measured with a stopwatch, is a dandy simplification. 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 last time on the stopwatch. When initial time is taken to exist zippo, nosotros apply 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 terminal values. That is, t is the final time, 10 is the last position, and v is the final velocity. This gives a simpler expression for elapsed fourth dimension, $\text{Δ}t=t$ . It also simplifies the expression for x displacement, which is now $\text{Δ}\mathrm{ten}=\mathrm{ten}-{\mathrm{10}}_0$ . Also, it simplifies the expression for modify in velocity, which is now $\text{Δ}5=v-{\mathrm{five}}_0$ . To summarize, using the simplified notation, with the initial fourth dimension taken to exist nix,

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

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

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

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

Thus, we can use the symbol a for acceleration at all times. Assuming acceleration to be abiding does not seriously limit the situations we can report nor does it degrade the accurateness of our treatment. For one thing, acceleration is constant in a great number of situations. Furthermore, in many other situations we tin can describe move accurately past bold a constant acceleration equal to the average acceleration for that motion. Lastly, for motion during which acceleration changes drastically, such as a machine accelerating to top speed and and so braking to a finish, move can be considered in separate parts, each of which has its own constant acceleration.

Displacement and Position from Velocity

To get our first 2 equations, nosotros start 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

$$\stackrel{\text{–}}{\mathrm{five}}=\frac{\mathrm{ten}-x_0}{t}.$$

Solving for x gives the states

where the boilerplate velocity is

The equation $\stackrel{\text{–}}{5}=\frac{v_0+v}{2}$ reflects the fact that when acceleration is constant, $\stackrel{\text{–}}{5}$ is just the simple boilerplate of the initial and final velocities. Effigy 3.18 illustrates this concept graphically. In part (a) of the effigy, acceleration is constant, with velocity increasing at a abiding rate. The boilerplate velocity during the ane-h interval from twoscore km/h to 80 km/h is 60 km/h:

$$\stackrel{\text{–}}{v}=\frac{5_0+5}{2}=\frac{40\text{km/h}+\mathrm{eighty}\text{km/h}}{\mathrm{ii}}=60\text{km/h}\text{.}$$

In part (b), acceleration is not abiding. During the ane-h interval, velocity is closer to eighty km/h than twoscore km/h. Thus, the average velocity is greater than in function (a).

Effigy iii.xviii (a) Velocity-versus-time graph with constant acceleration showing the initial and final velocities ${\mathrm{five}}_0\text{and}v$. The average velocity is $\frac{1}{2}({\mathrm{five}}_0+v)=\mathrm{sixty}\text{km}\text{/}\text{h}$. (b) Velocity-versus-time graph with an dispatch that changes with time. The average velocity is not given past $\frac{\mathrm{one}}{2}({\mathrm{five}}_0+v)$, but is greater than threescore km/h.

Solving for Concluding Velocity from Acceleration and Time

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

$$a=\frac{\text{Δ}\mathrm{five}}{\text{Δ}t}.$$

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

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

Solving for five yields

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

iii.12

Example three.7

Calculating Final Velocity

An airplane lands with an initial velocity of seventy.0 one thousand/s and so accelerates opposite to the motility at 1.fifty m/southwardⁱⁱ for 40.0 s. What is its final velocity?

Strategy

Starting time, we identify the knowns: $v_0=\mathrm{seventy}\text{m/s,}a=\mathrm{-i.50}{\text{m/south}}^2,t=40\text{s}$ .

Second, nosotros identify the unknown; in this case, information technology is final velocity ${\mathrm{five}}_{\text{f}}$ .

Terminal, we determine which equation to apply. To practise this nosotros figure out which kinematic equation gives the unknown in terms of the knowns. We summate the final velocity using Equation 3.12, $v=v_0+at$ .

Solution

Substitute the known values and solve:

$$v={\mathrm{five}}_0+at=70.0\text{m/s}+\left(\mathrm{-1.fifty}\text{1000/}{\text{s}}^2\right)\left(\mathrm{40.0\; s}\right)=\mathrm{10.0\; m/s.}$$

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

Effigy iii.19 The airplane lands with an initial velocity of 70.0 m/due south and slows to a final velocity of x.0 g/south earlier heading for the terminal. Note the dispatch is negative because its management is reverse to its velocity, which is positive.

Significance

The final velocity is much less than the initial velocity, every bit desired when slowing downward, but is however positive (encounter figure). With jet engines, reverse thrust can be maintained long enough to stop the aeroplane and start moving it astern, which is indicated by a negative final velocity, merely is non the instance here.

In add-on to existence useful in trouble solving, the equation $v=v_0+at$ gives us insight into the relationships among velocity, acceleration, and fourth dimension. Nosotros tin can see, for case, that

• Final velocity depends on how big the acceleration is and how long it lasts
• If the acceleration is nothing, and so the final velocity equals the initial velocity (v = v ₀), equally expected (in other words, velocity is constant)
• If a is negative, so the last velocity is less than the initial velocity

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

Solving for Last Position with Constant Acceleration

We tin can combine the previous equations to find a 3rd equation that allows us to calculate the terminal position of an object experiencing constant acceleration. We outset with

$$v=5_0+at.$$

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

$$\frac{5_0+5}{2}=5_0+\frac{1}{\mathrm{two}}at.$$

Since $\frac{{\mathrm{five}}_0+5}{2}=\stackrel{\text{–}}{\mathrm{five}}$ for constant dispatch, we have

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

At present nosotros substitute this expression for $\stackrel{\text{–}}{v}$ into the equation for displacement, $x=x_0+\stackrel{\text{–}}{v}t$ , yielding

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

3.13

Example three.8

Calculating Displacement of an Accelerating Object

Dragsters can achieve an average acceleration of 26.0 m/south². Suppose a dragster accelerates from rest at this rate for v.56 s Figure 3.xx. How far does information technology travel in this fourth dimension?

Figure 3.20 U.Due south. Army Meridian 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

First, let's describe a sketch Figure three.21. We are asked to detect deportation, which is x if we take $x_0$ to be zippo. (Retrieve about $x_0$ as the starting line of a race. Information technology tin can be anywhere, just we call it naught and measure all other positions relative to it.) We tin use the equation $x=x_0+5_{0}t+\frac{1}{\mathrm{ii}}a{t}^2$ when we identify $v_0$ , $a$ , and t from the argument of the problem.

Figure three.21 Sketch of an accelerating dragster.

Solution

First, we need to place the knowns. Starting from rest means that $v_0=0$ , a is given as 26.0 m/s² and t is given as 5.56 southward.

2d, we substitute the known values into the equation to solve for the unknown:

$$x={\mathrm{10}}_0+v_{0}t+\frac{\mathrm{i}}{2}a{t}^2.$$

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

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

Substituting the identified values of a and t gives

$$x=\frac{\mathrm{ane}}{2}(26.0{\text{thou/due south}}^{\mathrm{two}}){(5.56\text{s})}^2=402\text{k}\text{.}$$

Significance

If we convert 402 m to miles, we notice that the distance covered is very close to ane-quarter of a mile, the standard distance for elevate racing. So, our answer is reasonable. This is an impressive displacement to cover in only five.56 s, but top-notch dragsters can exercise a quarter mile in even less time than this. If the dragster were given an initial velocity, this would add another term to the altitude equation. If the same acceleration and time are used in the equation, the distance covered would exist much greater.

What else can we larn past examining the equation $x=x_0+5_{0}t+\frac{1}{2}a{t}^2?$ Nosotros can run across the following relationships:

• Deportation depends on the foursquare of the elapsed time when acceleration is non naught. In Example 3.viii, the dragster covers just one-fourth of the total distance in the first half of the elapsed time.
• If acceleration is cypher, then initial velocity equals boilerplate velocity $(v_0=\stackrel{\text{–}}{v})$, and $x={\mathrm{10}}_0+5_{0}t+\frac{\mathrm{i}}{2}a{t}^{\mathrm{ii}}\text{becomes}x={\mathrm{10}}_0+v_{0}t.$

Solving for Concluding Velocity from Altitude and Acceleration

A quaternary useful equation can be obtained from some other algebraic manipulation of previous equations. If we solve $v=v_0+at$ for t, we get

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

Substituting this and $\stackrel{\text{–}}{v}=\frac{v_0+v}{2}$ into $x={\mathrm{10}}_0+\stackrel{\text{–}}{v}t$ , we get

$$v^{\mathrm{ii}}={\mathrm{five}}_0^{\mathrm{two}}+2a(\mathrm{ten}-x_0)(\text{constant}a).$$

3.14

Example 3.9

Calculating Final Velocity

Calculate the final velocity of the dragster in Instance 3.viii without using information virtually time.

Strategy

The equation ${\mathrm{five}}^{\mathrm{two}}=v_0^2+2a(\mathrm{10}-x_0)$ is ideally suited to this task considering it relates velocities, dispatch, 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 remainder. Nosotros besides know that 10 − x ₀ = 402 m (this was the answer in Example three.eight). The average acceleration was given past a = 26.0 m/southward^two.

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

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

Thus,

$$\begin{array}{ccc}{v}^{\mathrm{ii}}\hfill & =\hfill & \mathrm{two.09}\times {10}^{\mathrm{four}}{\text{m}}^{\mathrm{ii}}{\text{/due south}}^2\hfill \\ \\ v\hfill & =\hfill & \sqrt{\mathrm{ii.09}\times {10}^{\mathrm{four}}{\text{grand}}^{\mathrm{two}}{\text{/south}}^{\mathrm{ii}}}=145\text{thousand/s}\text{.}\hfill \end{array}$$

Significance

A velocity of 145 grand/s is about 522 km/h, or about 324 mi/h, simply even this breakneck speed is short of the record for the quarter mile. Also, annotation that a square root has two values; nosotros took the positive value to betoken a velocity in the same direction every bit the acceleration.

An test of the equation $v^{\mathrm{ii}}=5_0^2+\mathrm{ii}a(\mathrm{10}-x_0)$ 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 information technology acts.
• For a stock-still acceleration, a car that is going twice equally fast doesn't simply stop in twice the distance. It takes much further to stop. (This is why we have reduced speed zones nigh schools.)

Putting Equations Together

In the following examples, we go on to explore one-dimensional movement, just in situations requiring slightly more algebraic manipulation. The examples also give insight into trouble-solving techniques. The note that follows is provided for easy reference to the equations needed. Exist enlightened that these equations are non contained. In many situations we have two unknowns and demand 2 equations from the prepare to solve for the unknowns. We need as many equations as in that location are unknowns to solve a given situation.

Summary of Kinematic Equations (constant a)

$$\mathrm{ten}=x_0+\stackrel{\text{–}}{5}t$$

$$\stackrel{\text{–}}{\mathrm{five}}=\frac{5_0+v}{2}$$

$$v=v_0+at$$

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

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

Before nosotros get into the examples, let'due south look at some of the equations more than closely to come across the behavior of acceleration at extreme values. Rearranging Equation iii.12, we accept

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

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

Similarly, rearranging Equation 3.14, we tin can express acceleration in terms of velocities and displacement:

$$a=\frac{5^{\mathrm{ii}}-v_0^2}{\mathrm{ii}(x-x_0)}.$$

Thus, for a finite difference between the initial and final velocities acceleration becomes space in the limit the displacement approaches zero. Acceleration approaches cypher in the limit the difference in initial and final velocities approaches zero for a finite displacement.

Instance 3.10

How Far Does a Motorcar Go?

On dry concrete, a car can advance reverse to the motility at a rate of 7.00 one thousand/due southⁱⁱ, whereas on wet concrete it tin advance opposite to the move at only 5.00 m/southward². Detect the distances necessary to terminate a car moving at 30.0 grand/southward (near 110 km/h) on (a) dry concrete and (b) wet concrete. (c) Repeat both calculations and find the displacement from the point where the commuter sees a traffic light plough red, taking into account his reaction time of 0.500 due south to get his foot on the brake.

Strategy

Beginning, nosotros need to depict a sketch Figure 3.22. To determine which equations are best to use, nosotros need to listing all the known values and identify exactly what we need to solve for.

Figure 3.22 Sample sketch to visualize dispatch opposite to the move and stopping distance of a motorcar.

Solution

1. Get-go, nosotros need to identify the knowns and what we want to solve for. We know that 5 ₀ = xxx.0 m/s, v = 0, and a = −7.00 thousand/s² (a is negative because information technology is in a direction opposite to velocity). Nosotros take x ₀ to be nothing. We are looking for deportation $\text{Δ}\mathrm{ten}$, or x − x ₀.
   Second, nosotros place the equation that will help us solve the problem. The best equation to use is

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

   This equation is best because information technology includes just one unknown, x. We know the values of all the other variables in this equation. (Other equations would allow united states to solve for x, but they require us to know the stopping time, t, which nosotros do not know. We could use them, only it would entail additional calculations.)
   Third, nosotros rearrange the equation to solve for x:

   $$\mathrm{ten}-{\mathrm{10}}_0=\frac{v^{\mathrm{ii}}-5_0^{\mathrm{two}}}{2a}$$

   and substitute the known values:

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

   Thus,

   $$x=64.3\text{m on dry concrete}\text{.}$$

2. This part can be solved in exactly the same manner as (a). The only difference is that the acceleration is −5.00 m/due south². The result is

   $$x_{\text{wet}}=90.0\text{m on moisture concrete.}$$

3. When the commuter reacts, the stopping distance is the same equally it is in (a) and (b) for dry out and wet concrete. So, to answer this question, we need to summate how far the automobile travels during the reaction time, so add together that to the stopping time. It is reasonable to assume the velocity remains constant during the commuter'south reaction fourth dimension.
   To do this, nosotros, over again, identify the knowns and what we want to solve for. We know that $\stackrel{\text{–}}{5}=\mathrm{xxx.0}\text{m/due south}$, $t_{\text{reaction}}=0.500\text{s}$, and $a_{\text{reaction}}=0$. We take $x_{\text{0-reaction}}$ to be zero. Nosotros are looking for ${\mathrm{10}}_{\text{reaction}}$.
   2d, as before, we identify the best equation to use. In this case, $\mathrm{10}=x_0+\stackrel{\text{–}}{v}t$ works well because the only unknown value is x, which is what we desire to solve for.
   Third, we substitute the knowns to solve the equation:

   $$x=0+\left(30.0\text{one thousand/s}\right)\left(0.500\text{s}\right)=\mathrm{fifteen}.0\text{m}.$$

   This ways the auto travels 15.0 yard while the commuter reacts, making the total displacements in the two cases of dry out and moisture physical 15.0 m greater than if he reacted instantly.
   Final, nosotros and so add the displacement during the reaction fourth dimension to the displacement when braking (Figure 3.23),

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

   and find (a) to be 64.iii k + 15.0 m = 79.3 m when dry and (b) to be 90.0 grand + 15.0 thou = 105 m when wet.

Effigy three.23 The distance necessary to stop a machine varies profoundly, depending on road 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. As well shown are the total distances traveled from the signal when the driver beginning sees a light turn red, assuming a 0.500-south reaction time.

Significance

The displacements found in this example seem reasonable for stopping a fast-moving motorcar. It should have longer to stop a machine on moisture pavement than dry. It is interesting that reaction time adds significantly to the displacements, simply more of import is the general arroyo to solving problems. We identify the knowns and the quantities to exist adamant, and then find an appropriate equation. If there is more than one unknown, we need as many independent equations as there are unknowns to solve. In that location is ofttimes more than i way to solve a trouble. The various parts of this instance can, in fact, be solved by other methods, only the solutions presented here are the shortest.

Instance three.eleven

Computing Fourth dimension

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

Strategy

First, nosotros draw a sketch Figure 3.24. We are asked to solve for fourth dimension t. As earlier, we place the known quantities to cull a user-friendly concrete relationship (that is, an equation with one unknown, t.)

Figure 3.24 Sketch of a machine accelerating on a state highway ramp.

Solution

Again, we identify the knowns and what nosotros want to solve for. We know that $x_0=0,$
$v_0=10\text{chiliad/s},a=2.00\text{m/}{\text{s}}^2$ , and 10 = 200 m.

We need to solve for t. The equation $\mathrm{ten}={\mathrm{10}}_0+{\mathrm{five}}_{0}t+\frac{1}{2}a{t}^2$ works best because the only unknown in the equation is the variable t, for which we need to solve. From this insight we come across that when nosotros input the knowns into the equation, we cease up with a quadratic equation.

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

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

We then simplify the equation. The units of meters cancel because they are in each term. We can go the units of seconds to abolish by taking t = t southward, where t is the magnitude of fourth dimension and s is the unit. Doing then leaves

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

We so 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 = ten.0 and t = −20.0. A negative value for time is unreasonable, since it would mean the event happened 20 s before the motion began. We can discard that solution. Thus,

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

Significance

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

Check Your Agreement iii.v

A rocket accelerates at a rate of xx 1000/s² during launch. How long does it take the rocket to reach a velocity of 400 one thousand/s?

Instance 3.12

Acceleration of a Spaceship

A spaceship has left Earth'due south orbit and is on its manner to the Moon. It accelerates at 20 g/s² for ii min and covers a distance of thou km. What are the initial and final velocities of the spaceship?

Strategy

We are asked to find the initial and final velocities of the spaceship. Looking at the kinematic equations, we see that i equation volition not give the answer. We must employ one kinematic equation to solve for one of the velocities and substitute it into another kinematic equation to get the second velocity. Thus, we solve 2 of the kinematic equations simultaneously.

Solution

First we solve for $5_0$ using $x=x_0+v_{0}t+\frac{1}{2}a{t}^2:$

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

$$\mathrm{i.0}\times {\mathrm{x}}^{\mathrm{vi}}\text{m}=v_0(120.0\text{due south})+\frac{1}{\mathrm{two}}(\mathrm{xx.0}{\text{thou/s}}^{\mathrm{two}}){(120.0\text{s})}^{\mathrm{two}}$$

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

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

$$\mathrm{five}=v_0+at=7133.3\text{m/due south}+(20.0{\text{m/s}}^{\mathrm{ii}})(120.0\text{due south})=\mathrm{9533.iii}\text{m/s.}$$

Significance

There are 6 variables in displacement, fourth dimension, velocity, and acceleration that draw motion in ane dimension. The initial conditions of a given problem tin be many combinations of these variables. Because of this diverseness, solutions may not be as easy every bit simple substitutions into one of the equations. This example illustrates that solutions to kinematics may require solving 2 simultaneous kinematic equations.

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

Two-Body Pursuit Issues

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

Effigy 3.25 A two-body pursuit scenario where car 2 has a constant velocity and machine 1 is behind with a constant acceleration. Automobile 1 catches up with motorcar two at a afterwards time.

The time and distance required for car 1 to catch car two depends on the initial distance car i is from automobile 2 as well as the velocities of both cars and the acceleration of automobile 1. The kinematic equations describing the motility of both cars must be solved to find these unknowns.

Consider the following example.

Example three.13

Chetah Catching a Gazelle

A cheetah waits in hiding backside a bush-league. The cheetah spots a gazelle running past at 10 yard/southward. At the instant the gazelle passes the cheetah, the cheetah accelerates from rest at 4 yard/due south² to catch the gazelle. (a) How long does information technology accept the cheetah to catch the gazelle? (b) What is the displacement of the gazelle and cheetah?

Strategy

We use the gear up of equations for constant acceleration to solve this problem. Since there are two objects in motion, nosotros have separate equations of motion describing each animate being. But what links the equations is a common parameter that has the same value for each beast. If nosotros look at the problem closely, it is clear the mutual parameter to each animal is their position x at a later fourth dimension t. Since they both start at $x_0=0$ , their displacements are the aforementioned at a later on time t, when the chetah catches up with the gazelle. If nosotros pick the equation of motion that solves for the displacement for each animal, nosotros tin can and then set the equations equal to each other and solve for the unknown, which is fourth dimension.

Solution

1. Equation for the gazelle: The gazelle has a constant velocity, which is its average velocity, since it is not accelerating. Therefore, nosotros use Equation 3.10 with ${\mathrm{ten}}_0=0$:

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

   Equation for the cheetah: The cheetah is accelerating from rest, then we use Equation 3.13 with ${\mathrm{10}}_0=0$ and $v_0=0$:

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

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

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

   The gazelle has a abiding velocity of 10 m/s, which is its average velocity. The acceleration of the cheetah is 4 thou/sⁱⁱ. Evaluating t, the fourth dimension for the cheetah to reach the gazelle, we have

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

2. To get the displacement, we use either the equation of motion for the cheetah or the gazelle, since they should both give the same answer.
   Displacement of the chetah:

   $$x=\frac{\mathrm{i}}{\mathrm{two}}a{t}^{\mathrm{two}}=\frac{1}{\mathrm{ii}}(4{\text{m/southward}}^{\mathrm{two}}){(\mathrm{five})}^2=50\text{chiliad}\text{.}$$

   Displacement of the gazelle:

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

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

Significance

It is important to clarify the motion of each object and to use the appropriate kinematic equations to describe the individual motion. It is too important to accept a good visual perspective of the ii-torso pursuit problem to see the common parameter that links the motion of both objects.

Check Your Understanding 3.half-dozen

A bicycle has a constant velocity of 10 m/south. A person starts from rest and begins to run to take hold of up to the bicycle in 30 due south when the bicycle is at the aforementioned position every bit the person. What is the acceleration of the person?

What Does the Scale Read (in N) When the Elevator Is Accelerating Upward at 2.3 M/s2 ?

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

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