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The Nature of Science and Physics

Kinematics

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Two-Dimensional Kinematics

Dynamics: Force and Newton's Laws of Motion

Introduction to Dynamics: Newton's Laws of MotionDevelopment of Force ConceptNewton's First Law of Motion: InertiaNewton's Second Law of Motion: Concept of a SystemNewton's Third Law of Motion: Symmetry in ForcesVideo: Newton's LawsNormal, Tension, and Other Examples of ForcesProblem-Solving StrategiesFurther Applications of Newton's Laws of MotionExtended Topic: The Four Basic Forces—An Introduction

Further Applications of Newton's Laws: Friction, Drag, and Elasticity

Uniform Circular Motion and Gravitation

Introduction to Uniform Circular Motion and GravitationRotation Angle and Angular VelocityCentripetal AccelerationCentripetal ForceFictitious Forces and Non-inertial Frames: The Coriolis ForceNewton's Universal Law of GravitationVideo: GravitationSatellites and Kepler's Laws: An Argument for Simplicity

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Linear Momentum and Collisions

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Physics of Hearing

- Calculate displacement of an object that is not accelerating, given initial position and velocity.
- Calculate final velocity of an accelerating object, given initial velocity, acceleration, and time.
- Calculate displacement and final position of an accelerating object, given initial position, initial velocity, time, and acceleration.

We might know that the greater the acceleration of, say, a car moving away from a stop sign, the greater the displacement in a given time. But we have not developed a specific equation that relates acceleration and displacement. In this section, we develop some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration already covered.

$\begin{cases}{\Delta}{t} &=& t \\{\Delta}{x} &=& x-{{x}_{0}}\\{\Delta}{v} &=& v-{{v}_{0}}\end{cases}\\$

where We now make the important assumption that

$\bar{a }=a=\text{ constant}\\$

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

Substituting the simplified notation for Δ*x* and Δ*t* yields

Solving for *x* yields

where the average velocity is

$\bar{v}=\frac{\Delta x}{\Delta t}\\$

.$\bar{v}=\frac{x-{x}_{0}}{t}\\$

$x={x}_{0}+\bar{v}t\\$

,$\bar{v}=\frac{{v}_{0}+v}{2}\left(\text{ constant }a\right)\\$

.
The equation *v* is just the simple average of the initial and final velocities. For example, if you steadily increase your velocity (that is, with constant acceleration) from 30 to 60 km/h, then your average velocity during this steady increase is 45 km/h. Using the equation

### Example 1. Calculating Displacement: How Far does the Jogger Run?

A jogger runs down a straight stretch of road with an average velocity of 4.00 m/s for 2.00 min. What is his final position, taking his initial position to be zero?

**Strategy**

Draw a sketch.

The final position *x* is given by the equation

To find *x*, we identify the values of *x*_{0}, *t* from the statement of the problem and substitute them into the equation.

**Solution**

**Discussion**

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

$\bar{v}=\frac{{v}_{0}+v}{2}\\$

reflects the fact that, when acceleration is constant, $\bar{v}=\frac{{v}_{0}+v}{2}\\$

to check this, we see that$\bar{v}=\frac{{v}_{0}+v}{2}=\frac{\text{30 km/h}+\text{60 km/h}}{2}=\text{45 km/h}\\$

,which seems logical.

$x={x}_{0}+\bar{v}t\\$

.$\bar{v}\\$

, and 1. Identify the knowns.

$\bar{v}=4.00\text{ m/s}\\$

, $\Delta t=2.00\text{ min}$

, and ${x}_{0}=0\text{ m}\\$

.
2. Enter the known values into the equation.

$x={x}_{0}+\bar{v}t=0+\left(4\text{.}\text{00 m/s}\right)\left(\text{120 s}\right)=\text{480 m}\\$

$x={x}_{0}+\bar{v}t\\$

gives insight into the relationship between displacement, average velocity, and time. It shows, for example, that displacement is a linear function of average velocity. (By linear function, we mean that displacement depends on $\bar{v}\\$

$\bar{v}\\$

raised to some other power, such as $\bar{v}^{2}\\$

. When graphed, linear functions look like straight lines with a constant slope.) On a car trip, for example, we will get twice as far in a given time if we average 90 km/h than if we average 45 km/h.
We can derive another useful equation by manipulating the definition of acceleration.

Substituting the simplified notation for Δ*v* and Δ*t* gives us

Solving for *v* yields

$a=\frac{\Delta v}{\Delta t}\\$

$a=\frac{v-{v}_{0}}{t}\text{}\left(\text{constant}a\right)\\$

$v={v}_{0}+\text{at}\text{}\left(\text{constant}a\right)\\$

2. Identify the unknown. In this case, it is final velocity,

3. Determine which equation to use. We can calculate the final velocity using the equation

$v={v}_{0}+{at}\\$

.4. Plug in the known values and solve.

$v={v}_{0}+\text{at}=\text{70}\text{.}\text{0 m/s}+\left(-1\text{.}{\text{50 m/s}}^{2}\right)\left(\text{40}\text{.}\text{0 s}\right)=\text{10}\text{.}\text{0 m/s}\\$

$v={v}_{0}+\text{at}\\$

gives us insight into the relationships among velocity, acceleration, and time. From it we can see, for example, that- final velocity depends on how large the acceleration is and how long it lasts
- if the acceleration is zero, then the final velocity equals the initial velocity (
*v*=*v*_{0}), as expected (i.e., velocity is constant) - if
is negative, then the final velocity is less than the initial velocity*a*

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

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

Adding *v*_{0} to each side of this equation and dividing by 2 gives

Since

Now we substitute this expression for

$v={v}_{0}+{at}\\$

$\frac{{v}_{0}+v}{2}={v}_{0}+\frac{1}{2}{at}\\$

$\frac{{v}_{0}+v}{2}=\bar{v}\\$

for constant acceleration, then$\bar{v}={v}_{0}+\frac{1}{2}{at}\\$

$\bar{v}\\$

into the equation for displacement, $x={x}_{0}+\bar{v}t\\$

, yielding$x={x}_{0}+{v}_{0}t+\frac{1}{2}{\text{at}}^{2}\left(\text{constant}a\right)\text{.}\\$

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

**Strategy**

Draw a sketch.

We are asked to find displacement, which is *x* if we take *x*_{0} to be zero. (Think about it like the starting line of a race. It can be anywhere, but we call it 0 and measure all other positions relative to it.) We can use the equation *v*_{0}, *a*, and *t* from the statement of the problem.

**Solution**

1. Identify the knowns. Starting from rest means that *v*_{0 }= 0, *a* is given as 26.0 m/s^{2} and *t* is given as 5.56 s.

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

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

Substituting the identified values of *a* and *t* gives

yielding

*x *= 402 m.
**Discussion**

If we convert 402 m to miles, we find that the distance covered is very close to one quarter of a mile, the standard distance for drag racing. So the answer is reasonable. This is an impressive displacement in only 5.56 s, but top-notch dragsters can do a quarter mile in even less time than this.

$x={x}_{0}+{v}_{0}t+\frac{1}{2}{{at}}^{2}\\$

once we identify 2. Plug the known values into the equation to solve for the unknown

$x={x}_{0}+{v}_{0}t+\frac{1}{2}{\text{at}}^{2}\text{.}\\$

$x=\frac{1}{2}{\text{at}}^{2}\text{.}\\$

$x=\frac{1}{2}\left(\text{26}\text{.}{\text{0 m/s}}^{2}\right){\left(5\text{.}\text{56 s}\right)}^{2}\\$

,$x={x}_{0}+{v}_{0}t+\frac{1}{2}{\text{at}}^{2}\\$

? We see that:- displacement depends on the square of the elapsed time when acceleration is not zero. In Example 3, the dragster covers only one fourth of the total distance in the first half of the elapsed time
- if acceleration is zero, then the initial velocity equals average velocity (${v}_{0}=\bar{v}\\$) and$x={x}_{0}+{v}_{0}t+\frac{1}{2}{\text{at}}^{2}\\$becomes
*x*=*x*_{0 }+*v*_{0}*t*

${v}^{2}={v}_{0}^{2}+2a\left(x-{x}_{0}\right)\\$

is ideally suited to this task because it relates velocities, acceleration, and displacement, and no time information is required.2. Plug the knowns into the equation

${v}^{2}={v}_{0}^{2}+2a\left(x-{x}_{0}\right)\\$

and solve for ${v}^{2}=2\text{.}\text{09}\times {\text{10}}^{4}{\text{m}}^{2}{\text{/s}}^{2}\\$

.$v=\sqrt{2\text{.}\text{09}\times {\text{10}}^{4}{\text{m}}^{2}{\text{/s}}^{2}}=\text{145 m/s}\\$

.${v}^{2}={v}_{0}^{2}+2a\left(x-{x}_{0}\right)\\$

can produce further 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 deceleration, a car that is going twice as fast doesn’t simply stop in twice the distance—it takes much further to stop. (This is why we have reduced speed zones near schools.)

$x={x}_{0}+\bar{v}t\\$

$\bar{v}=\frac{{v}_{0}+v}{2}\\$

$v={v}_{0}+\text{at}\\$

$x={x}_{0}+{v}_{0}t+\frac{1}{2}{\text{at}}^{2}\\$

${v}^{2}={v}_{0}^{2}+2a\left(x-{x}_{0}\right)\\$

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

${v}^{2}={v}_{0}^{2}+2a\left(x-{x}_{0}\right)\\$

.3. Rearrange the equation to solve for

$x-{x}_{0}=\frac{{v}^{2}-{v}_{0}^{2}}{2a}\\$

$x - 0=\frac{{0}^{2}-{\left(\text{30}\text{.}\text{0 m/s}\right)}^{2}}{2\left(-7\text{.}{\text{00 m/s}}^{2}\right)}\\$

1. Identify the knowns and what we want to solve for. We know that

$\bar{v}=30.0 \text{ m/s}\\$

; 2. Identify the best equation to use.

$x={x}_{0}+\bar{v}t\\$

works well because the only unknown value is 3. Plug in the knowns to solve the equation.

4. Add the displacement during the reaction time to the displacement when braking.

- 64.3 m + 15.0 m = 79.3 m when dry
- 90.0 m + 15.0 m = 105 m when wet

Suppose a car merges into freeway traffic on a 200-m-long ramp. If its initial velocity is 10.0 m/s and it accelerates at 2.00 m/s^{2}, how long does it take to travel the 200 m up the ramp? (Such information might be useful to a traffic engineer.)

**Strategy**

Draw a sketch.

We are asked to solve for the time *t*. As before, we identify the known quantities in order to choose a convenient physical relationship (that is, an equation with one unknown, *t*).

**Solution**

1. Identify the knowns and what we want to solve for. We know that *v*_{0 }= 10 m/s; *a *= 2.00 m/s^{2}; and *x *= 200 m.

2. We need to solve for*t*. Choose the best equation. *t* for which we need to solve.

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

*t* = *ts*, where *t* is the magnitude of time and s is the unit. Doing so leaves

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

*t*^{ 2 } + 10*t* − 200 = 0
This is a quadratic equation of the form

*a* = 1.00, *b* = 10.0 and *c* = -200.

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

This yields two solutions for *t*, which are

*t *= *t* in seconds, or

**Discussion**

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

2. We need to solve for

$x={x}_{0}+{v}_{0}t+\frac{1}{2}{{at}}^{2}\\$

works best because the only unknown in the equation is the variable 3. We will need to rearrange the equation to solve for

$\text{200 m}=\text{0 m}+\left(\text{10}\text{.}\text{0 m/s}\right)t+\frac{1}{2}\left(2\text{.}{\text{00 m/s}}^{2}\right){t}^{2}\\$

4. Simplify the equation. The units of meters (m) cancel because they are in each term. We can get the units of seconds (s) to cancel by taking 200 = 10*t *+ *t*^{2}.

5. Use the quadratic formula to solve for ${\text{at}}^{2}+\text{bt}+c=0\\$

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

$t=\frac{-b\pm \sqrt{{b}^{2}-4\text{ac}}}{2a}\\$

*t* = 10.0 and -20.0.

*t* = 10.0 s and -20.0 s.

*t* = 10.0 s.

We have been using SI units of meters per second squared to describe some examples of acceleration or deceleration of cars, runners, and trains. To achieve a better feel for these numbers, one can measure the braking deceleration of a car doing a slow (and safe) stop. Recall that, for average acceleration,

$\bar{a}=\Delta v/ \Delta t\\$

. While traveling in a car, slowly apply the brakes as you come up to a stop sign. Have a passenger note the initial speed in miles per hour and the time taken (in seconds) to stop. From this, calculate the deceleration in miles per hour per second. Convert this to meters per second squared and compare with other decelerations mentioned in this chapter. Calculate the distance traveled in braking.
A manned rocket accelerates at a rate of 20 m/s^{2} during launch. How long does it take the rocket reach a velocity of 400 m/s?

To answer this, choose an equation that allows you to solve for time *t*, given only *a*, *v*_{0}, and *v*.

*t**. *

$v={v}_{0}+{at}\\$

Rearrange to solve for $t=\frac{v-v{}_{0}\text{}}{a}=\frac{\text{400 m/s}-\text{0 m/s}}{{\text{20 m/s}}^{2}}=\text{20 s}\\$

- To simplify calculations we take acceleration to be constant, so that $\bar{a}=a\\$at all times.
- We also take initial time to be zero.
- Initial position and velocity are given a subscript 0; final values have no subscript. Thus,

$\begin{cases}{\Delta}{t} &=& t \\{\Delta}{x} &=& x-{{x}_{0}}\\{\Delta}{v} &=& v-{{v}_{0}}\end{cases}\\$ - The following kinematic equations for motion with constant $a$are useful:

$x={x}_{0}+\bar{v}t\\$$\bar{v}=\frac{{v}_{0}+v}{2}\\$$v={v}_{0}+\text{at}\\$$x={x}_{0}+{v}_{0}t+\frac{1}{2}{\text{at}}^{2}\\$In vertical motion,${v}^{2}={v}_{0}^{2}+2a\left(x-{x}_{0}\right)\\$*y*is substituted for*x*.

2. A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10 × 10

3. A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 6.20 × 10

4. (a) A light-rail commuter train accelerates at a rate of 1.35 m/s

5. While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s

6. At the end of a race, a runner decelerates from a velocity of 9.00 m/s at a rate of 2.00 m/s

7.

8. In a slap shot, a hockey player accelerates the puck from a velocity of 8.00 m/s to 40.0 m/s in the same direction. If this shot takes 3.33 × 10

9. A powerful motorcycle can accelerate from rest to 26.8 m/s (100 km/h) in only 3.90 s. (a) What is its average acceleration? (b) How far does it travel in that time?

10. Freight trains can produce only relatively small accelerations and decelerations. (a) What is the final velocity of a freight train that accelerates at a rate of 0.0500 m/s

11. A fireworks shell is accelerated from rest to a velocity of 65.0 m/s over a distance of 0.250 m. (a) How long did the acceleration last? (b) Calculate the acceleration.

12. A swan on a lake gets airborne by flapping its wings and running on top of the water. (a) If the swan must reach a velocity of 6.00 m/s to take off and it accelerates from rest at an average rate of 0.350 m/s

13.

14. An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350 m. (a) What is his deceleration? (b) How long does the collision last?

15. In World War II, there were several reported cases of airmen who jumped from their flaming airplanes with no parachute to escape certain death. Some fell about 20,000 feet (6000 m), and some of them survived, with few life-threatening injuries. For these lucky pilots, the tree branches and snow drifts on the ground allowed their deceleration to be relatively small. If we assume that a pilot’s speed upon impact was 123 mph (54 m/s), then what was his deceleration? Assume that the trees and snow stopped him over a distance of 3.0 m.

16. Consider a grey squirrel falling out of a tree to the ground. (a) If we ignore air resistance in this case (only for the sake of this problem), determine a squirrel’s velocity just before hitting the ground, assuming it fell from a height of 3.0 m. (b) If the squirrel stops in a distance of 2.0 cm through bending its limbs, compare its deceleration with that of the airman in the previous problem.

17. An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of

$0\text{.}{\text{150 m/s}}^{2}$

as it goes through. The station is 210 m long. (a) How long is the nose of the train in the station? (b) How fast is it going when the nose leaves the station? (c) If the train is 130 m long, when does the end of the train leave the station? (d) What is the velocity of the end of the train as it leaves?18. Dragsters can actually reach a top speed of 145 m/s in only 4.45 s—considerably less time than given in Example 2.10 and Example 2.11. (a) Calculate the average acceleration for such a dragster. (b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402 m (a quarter mile) without using any information on time. (c) Why is the final velocity greater than that used to find the average acceleration? Hint: Consider whether the assumption of constant acceleration is valid for a dragster. If not, discuss whether the acceleration would be greater at the beginning or end of the run and what effect that would have on the final velocity.

19. A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates at the rate of 0.500 m/s

20. In 1967, New Zealander Burt Munro set the world record for an Indian motorcycle, on the Bonneville Salt Flats in Utah, with a maximum speed of 183.58 mi/h. The one-way course was 5.00 mi long. Acceleration rates are often described by the time it takes to reach 60.0 mi/h from rest. If this time was 4.00 s, and Burt accelerated at this rate until he reached his maximum speed, how long did it take Burt to complete the course?

21. (a) A world record was set for the men’s 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt "coasted" across the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration. (b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race?

(b) 2. 38.9 m/s (about 87 miles per hour)

4. (a) 16.5 s (b) 13.5 s (c) -2.68 m/s

6. (a) 20.0 m (b) -1.00 m/s (c) This result does not really make sense. If the runner starts at 9.00 m/s and decelerates at 2.00 m/s

8. 0.799 m

10. (a) 28.0 m/s (b) 50.9 s (c) 7.68 km to accelerate and 713 m to decelerate

12. (a) 51.4 m (b) 17.1 s

14. (a) -80 m/s

16. (a) 7.7 m/s (b) -15 × 10

18. (a) 36.2 m/s

20. 104 s

21. (a)