Rectilinear Translation (Motion Along a Straight Line)

Motion with constant velocity (uniform motion)

Motion with constant acceleration

Free-falling body

Note: From motion with constant acceleration, set vi = 0, vf = v, s = h, and a = g to get the free-fall formulas.

Motion with variable acceleration

Where = distance = height = velocity = initial velocity = final velocity

= acceleration = acceleration due to gravity (g = 9.81 m/s2 in SI = 32.2

ft/s2 in English) = time

Note:• is positive (+) if is increasing (accelerate).• is negative (-) if is decreasing (decelerate).• is positive (+) if the particle is moving downward.• is negative (-)if the particle is moving upward.

Useful conversion factors:

From To Multiply by

Kilometers per hour (kph) Meter per second (m/sec)

Meter per second (m/sec) Kilometers per hour (kph or km/hr)

Miles per hour (mph) Feet per second (fps or ft/sec)

Feet per second (ft/sec) Miles per hour (mph or mi/hr)

1002 Location of warning torpedo | Rectilinear Translation

Problem 1002On a certain stretch of track, trains run at 60 mph (96.56 kph). How far back of a stopped train should be a warning torpedo be placed to signal an oncoming train? Assume that the brakes are applied at once and retard the train at the uniform rate of 2 ft/sec2 (0.61 m/s2). Solution

English System

Initial velocity

SI units

Initial velocity

1003 Return in 10 seconds | Rectilinear Translation

Problem 1003A stone is thrown vertically upward and return to earth in 10 sec. What was its initial velocity and how high

did it go?

Solution

Return in 10 seconds = 5 seconds upward + 5 seconds downward

SI Units

Going upward (velocity at the highest point is zero):

answer

Going downward (initial velocity is zero; free-fall):

answer

English System

Going upward (velocity at the highest point is zero):

answer

Going downward (initial velocity is zero; free-fall):

answer

1007 Finding the time and passing point of stones | Rectilinear Translation

Problem 1007A stone is dropped from a captive balloon at an elevation of 1000 ft (304.8 m). Two seconds later another stone is thrown vertically upward from the ground with a velocity of 248 ft/s (75.6 m/s). If g = 32 ft/s2 (9.75 m/s2), when and where the stone pass each other?

Solution in English Units

Stone dropped from captive balloon (free-falling body):

Stone thrown vertically from the ground 2 seconds later

The stones will pass each other 5 seconds after the first stone was dropped from the captive balloon. answer

The stones will meet at a point 600 ft above the ground. answer

Solution in SI Units

Stone dropped from captive balloon (free-falling body):

Stone thrown vertically from the ground 2 seconds later

The stones will pass each other 5 seconds after the first stone was dropped from the captive balloon. answer

The stones will meet at a point 182.925 m above the ground. Answer

Curvilinear Translation (Projectile Motion)

Projectile motion follows a parabolic trajectory. The vertical component of projectile is under constant gravitational acceleration and the horizontal component is at constant velocity. For easy handling, resolve the motion into x and y components and use the formulas in rectilinear translation.

Form the figure below:

At any point B

For the x-component of motion, acceleration is zero (constant velocity), thus ax = 0.

For the y-component of motion, ay = -g. Notice that the first three formulas that follow are taken from motion with constant acceleration.

From x = voxt, t = x/vox. Substitute t = x/vox to y = voyt - ½ gt2.

At point A

At the highest point or summit, vAy = 0.

At point C

x = R, y = 0, vC = vo, and vy = -voy

Note:

vy is positive if directed upward and negative if directed downward

At any point D below the origin O, the sign of y is negative.

1008 Stones thrown vertically upward | Rectilinear Translation

Problem 1008A stone is thrown vertically upward from the ground with a velocity of 48.3 ft per sec (14.72 m per sec). One second later another stone is thrown vertically upward with a velocity of 96.6 ft per sec (29.44 m per sec). How far above the ground will the stones be at the same level?

Solution in English System

For the first stone:

For the second stone

answer

Solution in SI

For the first stone:

For the second stone

answer

In the above equations, the symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stands for the acceleration of the object. And the symbol v stands for the instantaneous velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value.

In this part of Lesson 6 we will investigate the process of using the equations to determine unknown information about an object's motion. The process involves the use of a problem-solving strategy that will be used throughout the course. The strategy involves the following steps:

a. Construct an informative diagram of the physical situation.

b. Identify and list the given information in variable form.

c. Identify and list the unknown information in variable form.

d. Identify and list the equation that will be used to determine unknown information from known information.

e. Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown information.

f. Check your answer to insure that it is reasonable and mathematically correct.

The use of this problem-solving strategy in the solution of the following problem is modeled in Examples A and B below.

Example A

Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s2, then determine the displacement of the car during the skidding process. (Note that the direction of the velocity and the acceleration vectors are denoted by a + and a - sign.)

The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. Note that the vf value can be inferred to be 0 m/s since Ima's car comes to a stop. The initial velocity (vi) of the car is +30.0 m/s since this is the velocity at the beginning of the motion (the skidding motion). And the acceleration (a) of the car is given as - 8.00 m/s2. (Always pay careful attention to the + and - signs for the given quantities.) The next step of thestrategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown quantity. The results of the first three steps are shown in the table below.

Diagram: Given:

vi = +30.0 m/svf = 0 m/s

a = - 8.00 m/s2

The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vf, vi, a, and d. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top right contains all four variables.

Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below.

(0 m/s)2 = (30.0 m/s)2 + 2*(-8.00 m/s2)*d

0 m2/s2 = 900 m2/s2 + (-16.0 m/s2)*d

(16.0 m/s2)*d = 900 m2/s2 - 0 m2/s2

(16.0 m/s2)*d = 900 m2/s2

d = (900 m2/s2)/ (16.0 m/s2)

d = (900 m2/s2)/ (16.0 m/s2)

d = 56.3 m

The solution above reveals that the car will skid a distance of 56.3 meters. (Note that this value is rounded to the third digit.)

The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. It takes a car a considerable distance to skid from 30.0 m/s (approximately 65 mi/hr) to a stop. The calculated distance is approximately one-half a football field, making this a very reasonable skidding distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is!

Example B

Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6.00 m/s2 for a time of 4.10 seconds. Determine the displacement of Ben's car during this time period.

Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step of the strategy involves the identification and listing of known information in variable form. Note that the vi value can be inferred to be 0 m/s since Ben's car is initially at rest. The acceleration (a) of the car is 6.00 m/s2. And the time (t) is given as 4.10 s. The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown information. The results of the first three steps are shown in the table below.

Diagram: Given:

vi = 0 m/st = 4.10 s

a = 6.00 m/s2

The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are t, vi, a, and d. An inspection of the four equations above reveals that the equation on the top left contains all four variables.

Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below.

d = (0 m/s)*(4.1 s) + 0.5*(6.00 m/s2)*(4.10 s)2

d = (0 m) + 0.5*(6.00 m/s2)*(16.81 s2)

d = 0 m + 50.43 m

d = 50.4 m

The solution above reveals that the car will travel a distance of 50.4 meters. (Note that this value is rounded to the third digit.)

The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. A car with an acceleration of 6.00 m/s/s will reach a speed of approximately 24 m/s (approximately 50 mi/hr) in 4.10 s. The distance over which such a car would be displaced during this time period would be approximately one-half a football field, making this a very reasonable distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is!

The two example problems above illustrate how the kinematic equations can be combined with a simple problem-solving strategy to predict unknown motion parameters for a moving object. Provided that three motion parameters are known, any of the remaining values can be determined. In

the next part of Lesson 6, we will see how this strategy can be applied to free fall situations. Or if interested, you can try some practice problems and check your answer against the given solutions.

Kinematic Equations and Free Fall

As mentioned in Lesson 5, a free-falling object is an object that is falling under the sole influence of gravity. That is to say that any object that is moving and being acted upon only be the force of gravity is said to be "in a state of free fall." Such an object will experience a downward acceleration of 9.8 m/s/s. Whether the object is falling downward or rising upward towards its peak, if it is under the sole influence of gravity, then its acceleration value is 9.8 m/s/s.

Like any moving object, the motion of an object in free fall can be described by four kinematic equations. The kinematic equations that describe any object's motion are:

The symbols in the above equation have a specific meaning: the symbol d stands for the displacement; the symbol tstands for the time; the symbol a stands for the acceleration of the object; the symbol vi stands for the initial velocity value; and the symbol vf stands for the final velocity.

There are a few conceptual characteristics of free fall motion that will be of value when using the equations to analyze free fall motion. These concepts are described as follows:

An object in free fall experiences an acceleration of -9.8 m/s/s. (The - sign indicates a downward acceleration.) Whether explicitly stated or not, the value of the acceleration in the kinematic equations is -9.8 m/s/s for any freely falling object.

If an object is merely dropped (as opposed to being thrown) from an elevated height, then the initial velocity of the object is 0 m/s.

If an object is projected upwards in a perfectly vertical direction, then it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value can be used as one of the motion parameters in the kinematic equations; for example, the final velocity (vf) after traveling to the peak would be assigned a value of 0 m/s.

If an object is projected upwards in a perfectly vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity that it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have a downward velocity of -30 m/s when it returns to the same height.

These four principles and the four kinematic equations can be combined to solve problems involving the motion of free falling objects. The two examples below illustrate application of free fall principles to kinematic problem-solving. In each example, the problem solving strategy that was introduced earlier in this lesson will be utilized.

Example A

Luke Autbeloe drops a pile of roof shingles from the top of a roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground.

The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 8.52 meters. The displacement (d) of the shingles is -8.52 m. (The - sign indicates that the displacement is downward). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. For example, the vi value can be inferred to be 0 m/s since the shingles are dropped (released from rest;see note above). And the acceleration (a) of the shingles can be inferred to be -9.8 m/s2 since the shingles are free-falling (see note above). (Always pay careful attention to the + and - signs for the given quantities.) The next step of the solution involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the time of

fall. So t is the unknown quantity. The results of the first three steps are shown in the table below.

Diagram: Given:

vi = 0.0 m/sd = -8.52 m

a = - 9.8 m/s2

The next step involves identifying a kinematic equation that allows you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are d, vi, a, and t. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top left contains all four variables.

Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below.

-8.52 m = (0 m/s)*(t) + 0.5*(-9.8 m/s2)*(t)2

-8.52 m = (0 m) *(t) + (-4.9 m/s2)*(t)2

-8.52 m = (-4.9 m/s2)*(t)2

(-8.52 m)/(-4.9 m/s2) = t2

1.739 s2 = t2

t = 1.32 s

The solution above reveals that the shingles will fall for a time of 1.32 seconds before hitting the ground. (Note that this value is rounded to the third digit.)

The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The shingles are falling a distance of approximately 10 yards (1 meter is pretty close to 1 yard); it seems that an answer between 1 and 2 seconds would be highly reasonable. The calculated time easily falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for time and insuring that the left side of

the equation is equal to the right side of the equation. Indeed it is!

Example B

Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height.

Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 26.2 m/s. The initial velocity (vi) of the vase is +26.2 m/s. (The + sign indicates that the initial velocity is an upwards velocity). The remaining information must be extracted from the problem statement based upon your understanding ofthe above principles. Note that the vf value can be inferred to be 0 m/s since the final state of the vase is the peak of its trajectory (see note above). The acceleration (a) of the vase is -9.8 m/s2 (see note above). The next step involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the vase (the height to which it rises above its starting height). So d is the unknown information. The results of the first three steps are shown in the table below.

Diagram: Given:

vi = 26.2 m/svf = 0 m/s

a = -9.8 m/s2

The next step involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vi, vf, a, and d. An inspection of the four equations above reveals that the equation on the top right contains all four variables.

Once the equation is identified and written down, the next step involves substituting known values into the equation and

using proper algebraic steps to solve for the unknown information. This step is shown below.

(0 m/s)2 = (26.2 m/s)2 + 2*(-9.8m/s2)*d

0 m2/s2 = 686.44 m2/s2 + (-19.6 m/s2)*d

(-19.6 m/s2)*d = 0 m2/s2 -686.44 m2/s2

(-19.6 m/s2)*d = -686.44 m2/s2

d = (-686.44 m2/s2)/ (-19.6 m/s2)

d = 35.0 m

The solution above reveals that the vase will travel upwards for a displacement of 35.0 meters before reaching its peak. (Note that this value is rounded to the third digit.)

The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The vase is thrown with a speed of approximately 50 mi/hr (merely approximate 1 m/s to be equivalent to 2 mi/hr). Such a throw will never make it further than one football field in height (approximately 100 m), yet will surely make it past the 10-yard line (approximately 10 meters). The calculated answer certainly falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is!

Kinematic equations provide a useful means of determining the value of an unknown motion parameter if three motion parameters are known. In the case of a free-fall motion, the acceleration is often known. And in many cases, another motion parameter can be inferred through a solid knowledge of some basic kinematic principles. The next part of Lesson 6 provides a wealth of practice problems with answers and solutions.

Check Your Understanding

a. An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.

See solution below.

b. A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.

See solution below.

c. Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.6 seconds, what will be his final velocity and how far will he fall?

See solution below.

d. A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled.

See solution below.

e. A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon.

See solution below.

f. Rocket-powered sleds are used to test the human response to acceleration. If a rocket-powered sled is accelerated to a speed of 444 m/s in 1.8 seconds, then what is the acceleration and what is the distance that the sled travels?

See solution below.

g. A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m. Determine the acceleration of the bike.

See solution below.

h. An engineer is designing the runway for an airport. Of the planes that will use the airport, the lowest acceleration rate is likely to be 3 m/s2. The takeoff speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is the minimum allowed length for the runway?

See solution below.

i. A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding distance of the car (assume uniform acceleration).

See solution below.

j. A kangaroo is capable of jumping to a height of 2.62 m. Determine the takeoff speed of the kangaroo.

See solution below.

k. If Michael Jordan has a vertical leap of 1.29 m, then what is his takeoff speed and his hang time (total time to move upwards to the peak and then return to the ground)?

See solution below.

l. A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance

of 0.840 m. Determine the acceleration of the bullet (assume a uniform acceleration).

See solution below.

m. A baseball is popped straight up into the air and has a hang-time of 6.25 s. Determine the height to which the ball rises before it reaches its peak. (Hint: the time to rise to the peak is one-half the total hang-time.)

See solution below.

n. The observation deck of tall skyscraper 370 m above the street. Determine the time required for a penny to free fall from the deck to the street below.

See solution below.

o. A bullet is moving at a speed of 367 m/s when it embeds into a lump of moist clay. The bullet penetrates for a distance of 0.0621 m. Determine the acceleration of the bullet while moving into the clay. (Assume a uniform acceleration.)

See solution below.

p. A stone is dropped into a deep well and is heard to hit the water 3.41 s after being dropped. Determine the depth of the well.

See solution below.

q. It was once recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the Jaguar skidded to a stop with a constant acceleration of -3.90 m/s2, determine the speed of the Jaguar before it began to skid.

See solution below.

r. A plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed. Determine the acceleration of the plane and the time required to reach this speed.

See solution below.

s. A dragster accelerates to a speed of 112 m/s over a distance of 398 m. Determine the acceleration (assume uniform) of the dragster.

See solution below.

t. With what speed in miles/hr (1 m/s = 2.23 mi/hr) must an object be thrown to reach a height of 91.5 m (equivalent to one football field)? Assume negligible air resistance.

Solutions to Above Problems

a.

Given:

a = +3.2 m/s2 t = 32.8 s vi = 0 m/s

Find:

d = ??

b. d = vi*t + 0.5*a*t2

c. d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s2)*(32.8 s)2

d. d = 1720 me. Return to Problem 1

f.  

g.

Given:

d = 110 m t = 5.21 s vi = 0 m/s

Find:

a = ??

h. d = vi*t + 0.5*a*t2

i. 110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s)2

j. 110 m = (13.57 s2)*ak. a = (110 m)/(13.57 s2)

l. a = 8.10 m/ s2

m. Return to Problem 2 n.  

o.

Given:

a = -9.8 m t = 2.6 s vi = 0 m/s

Find:

d = ??

vf = ??

p. d = vi*t + 0.5*a*t2

q. d = (0 m/s)*(2.6 s)+ 0.5*(-9.8 m/s2)*(2.6 s)2

r. d = -33 m (- indicates direction)s. vf = vi + a*t

t. vf = 0 + (-9.8 m/s2)*(2.6 s)u. vf = -25.5 m/s (- indicates direction)

v. Return to Problem 3 w.  

x.

Given:

vi = 18.5 m/s vf = 46.1 m/s t = 2.47 s

Find:

d = ??

a = ??

y. a = (Delta v)/t

z. a = (46.1 m/s - 18.5 m/s)/(2.47 s)aa. a = 11.2 m/s2

bb. d = vi*t + 0.5*a*t2

cc. d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s2)*(2.47 s)2

dd. d = 45.7 m + 34.1 mee. d = 79.8 m

ff. (Note: the d can also be calculated using the equation vf

2 = vi2 + 2*a*d)

gg. Return to Problem 4 hh.  

ii.

Given:

vi = 0 m/s d = -1.40 m a = -1.67 m/s2

Find:

t = ??

jj. d = vi*t + 0.5*a*t2

kk. -1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s2)*(t)2

ll. -1.40 m = 0+ (-0.835 m/s2)*(t)2

mm. (-1.40 m)/(-0.835 m/s2) = t2

nn. 1.68 s2 = t2

oo. t = 1.29 spp. Return to Problem 5

qq.  

rr.

Given:

vi = 0 m/s vf = 44 m/s t = 1.80 s

Find:

a = ??

d = ??

ss. a = (Delta v)/t

tt. a = (444 m/s - 0 m/s)/(1.80 s)uu. a = 247 m/s2

vv. d = vi*t + 0.5*a*t2

ww. d = (0 m/s)*(1.80 s)+ 0.5*(247 m/s2)*(1.80 s)2

xx. d = 0 m + 400 myy. d = 400 m

zz. (Note: the d can also be calculated using the equation vf

2 = vi2 + 2*a*d)

aaa. Return to Problem 6 bbb.  

ccc.

Given:

vi = 0 m/s vf = 7.10 m/s d = 35.4 m

Find:

a = ??

ddd. vf2 = vi

2 + 2*a*d

eee. (7.10 m/s)2 = (0 m/s)2 + 2*(a)*(35.4 m)

fff. 50.4 m2/s2 = (0 m/s)2 + (70.8 m)*aggg. (50.4 m2/s2)/(70.8 m) = a

hhh. a = 0.712 m/s2

iii. Return to Problem 7 jjj.  

kkk.

Given:

vi = 0 m/s vf = 65 m/s a = 3 m/s2

Find:

d = ??

lll. vf2 = vi

2 + 2*a*d

mmm. (65 m/s)2 = (0 m/s)2 + 2*(3 m/s2)*d

nnn. 4225 m2/s2 = (0 m/s)2 + (6 m/s2)*dooo. (4225 m2/s2)/(6 m/s2) = d

ppp. d = 704 mqqq. Return to Problem 8

rrr.  

sss.

Given:

vi = 22.4 m/s vf = 0 m/s t = 2.55 s

Find:

d = ??

ttt. d = (vi + vf)/2 *t

uuu. d = (22.4 m/s + 0 m/s)/2 *2.55 svvv. d = (11.2 m/s)*2.55 s

www. d = 28.6 mxxx. Return to Problem 9

yyy.  

zzz.

Given:

a = -9.8 m/s2 vf = 0 m/s d = 2.62 m

Find:

vi = ??

aaaa. vf2 = vi

2 + 2*a*d

bbbb. (0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(2.62 m)

cccc. 0 m2/s2 = vi2 - 51.35 m2/s2

dddd. 51.35 m2/s2 = vi2

eeee. vi = 7.17 m/sffff. Return to Problem 10

gggg.  

hhhh.

Given:

a = -9.8 m/s2 vf = 0 m/s d = 1.29 m

Find:

vi = ??

t = ??

iiii. vf2 = vi

2 + 2*a*d

jjjj. (0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(1.29 m)

kkkk. 0 m2/s2 = vi2 - 25.28 m2/s2

llll. 25.28 m2/s2 = vi2

mmmm. vi = 5.03 m/snnnn. To find hang time, find the time to

the peak and then double it.oooo. vf = vi + a*t

pppp. 0 m/s = 5.03 m/s + (-9.8 m/s2)*tup

qqqq. -5.03 m/s = (-9.8 m/s2)*tup

rrrr. (-5.03 m/s)/(-9.8 m/s2) = tup

ssss. tup = 0.513 stttt. hang time = 1.03 s

uuuu. Return to Problem 11 vvvv.  

wwww.

Given:

vi = 0 m/s vf = 521 m/s d = 0.840 m

Find:

a = ??

xxxx. vf2 = vi

2 + 2*a*d

yyyy. (521 m/s)2 = (0 m/s)2 + 2*(a)*(0.840 m)

zzzz. 271441 m2/s2 = (0 m/s)2 + (1.68 m)*a

aaaaa. (271441 m2/s2)/(1.68 m) = abbbbb. a = 1.62*105 m /s2

ccccc. Return to Problem 12 ddddd.  

eeeee.

Given:

a = -9.8 m/s2 vf = 0 m/s t = 3.13 s

Find:

d = ??

a. (NOTE: the time required to move to the peak of the

trajectory is one-half the total hang time.)

First use: vf = vi + a*t

0 m/s = vi + (-9.8 m/s2)*(3.13 s)

0 m/s = vi - 30.6 m/s

vi = 30.6 m/s

Now use: vf2 = vi

2 + 2*a*d

(0 m/s)2 = (30.6 m/s)2 + 2*(-9.8 m/s2)*(d)

0 m2/s2 = (938 m/s) + (-19.6 m/s2)*d

-938 m/s = (-19.6 m/s2)*d

(-938 m/s)/(-19.6 m/s2) = d

d = 47.9 m

Return to Problem 13

 

fffff.

Given:

vi = 0 m/s d = -370 m a = -9.8 m/s2

Find:

t = ??

ggggg. d = vi*t + 0.5*a*t2

hhhhh. -370 m = (0 m/s)*(t)+ 0.5*(-9.8 m/s2)*(t)2

iiiii. -370 m = 0+ (-4.9 m/s2)*(t)2

jjjjj. (-370 m)/(-4.9 m/s2) = t2

kkkkk. 75.5 s2 = t2

lllll. t = 8.69 smmmmm. Return to Problem 14

nnnnn.  

ooooo.

Given:

vi = 367 m/s vf = 0 m/s d = 0.0621 m

Find:

a = ??

ppppp. vf2 = vi

2 + 2*a*d

qqqqq. (0 m/s)2 = (367 m/s)2 + 2*(a)*(0.0621 m)

rrrrr. 0 m2/s2 = (134689 m2/s2) + (0.1242 m)*a

sssss. -134689 m2/s2 = (0.1242 m)*attttt. (-134689 m2/s2)/(0.1242 m) = a

uuuuu. a = -1.08*106 m /s2

vvvvv. (The - sign indicates that the bullet slowed down.)

wwwww. Return to Problem 15 xxxxx.  

yyyyy.

Given:

a = -9.8 m/s2 t = 3.41 s vi = 0 m/s

Find:

d = ??

zzzzz. d = vi*t + 0.5*a*t2

aaaaaa.d = (0 m/s)*(3.41 s)+ 0.5*(-9.8 m/s2)*(3.41 s)2

bbbbbb. d = 0 m+ 0.5*(-9.8 m/s2)*(11.63 s2)

cccccc. d = -57.0 mdddddd. (NOTE: the - sign indicates

direction)eeeeee. Return to Problem 16

ffffff.  

gggggg.

Given:

a = -3.90 m/s2 vf = 0 m/s d = 290 m

Find:

vi = ??

hhhhhh. vf2 = vi

2 + 2*a*d

iiiiii. (0 m/s)2 = vi2 + 2*(-3.90 m/s2)*(290 m)

jjjjjj. 0 m2/s2 = vi2 - 2262 m2/s2

kkkkkk. 2262 m2/s2 = vi2

llllll. vi = 47.6 m /smmmmmm. Return to Problem 17

nnnnnn.  

oooooo.

Given:

vi = 0 m/s vf = 88.3 m/s d = 1365 m

Find:

a = ??

t = ??

pppppp. vf2 = vi

2 + 2*a*d

qqqqqq. (88.3 m/s)2 = (0 m/s)2 + 2*(a)*(1365 m)

rrrrrr. 7797 m2/s2 = (0 m2/s2) + (2730 m)*assssss. 7797 m2/s2 = (2730 m)*atttttt. (7797 m2/s2)/(2730 m) = a

uuuuuu. a = 2.86 m/s2

vvvvvv.vf = vi + a*twwwwww. 88.3 m/s = 0 m/s + (2.86

m/s2)*txxxxxx. (88.3 m/s)/(2.86 m/s2) = t

yyyyyy. t = 30. 8 szzzzzz. Return to Problem 18

aaaaaaa.  

bbbbbbb.

Given:

vi = 0 m/s vf = 112 m/s d = 398 m

Find:

a = ??

ccccccc. vf2 = vi

2 + 2*a*d

ddddddd. (112 m/s)2 = (0 m/s)2 + 2*(a)*(398 m)

eeeeeee. 12544 m2/s2 = 0 m2/s2 + (796 m)*a

fffffff. 12544 m2/s2 = (796 m)*aggggggg. (12544 m2/s2)/(796 m) = a

hhhhhhh. a = 15.8 m/s2

iiiiiii. Return to Problem 19 jjjjjjj.  

kkkkkkk.

Given:

a = -9.8 m/s2 vf = 0 m/s d = 91.5 m

Find:

vi = ??

t = ??

lllllll. First, find speed in units of m/s:

mmmmmmm. vf2 = vi

2 + 2*a*dnnnnnnn. (0 m/s)2 = vi

2 + 2*(-9.8 m/s2)*(91.5 m)

ooooooo. 0 m2/s2 = vi2 - 1793 m2/s2

ppppppp. 1793 m2/s2 = vi2

qqqqqqq. vi = 42.3 m/srrrrrrr. Now convert from m/s to mi/hr:

sssssss.vi = 42.3 m/s * (2.23 mi/hr)/(1 m/s)ttttttt. vi = 94.4 mi/hr