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- Introduction and Examples: Speed / Average Velocity / Acceleration
- A. Average Velocity, Average Speed Problems
- B. Constant Acceleration Problems
- C. Two-Object Problems
- D. Other Problems

**Speed** is scalar. Scalars are quantities with only magnitude. The direction does not matter. If you are on the highway whether traveling 100 km/h south or 100 km/h north, your speed is still 100 km/h. Other examples of scalar quantities are shoe size, mass, area, energy.

**(average speed) = (total distance)÷(total time)**

If someone walked 400 m in a straight line in 5 min, their average speed would be (400 m)÷(5 min) = 80 m/min. If the same person walked 100 m [North] then 300 m [South] in 5 minutes, their average speed would still be (400÷5) = 80 m/min. If that person walked 100 m [E] in .75 min, 100 m [N] in 1.50 min, 100 m [W] in 1.00 min and finally 100 m [S] in 1.75 min, their average speed would be (400 m)÷(5 min) = 80 m/min.

You drive a car for 2.0 h at 40 km/h, then for another 2.0 h at 60 km/h. a. What is your average speed? b. Do you get the same answer if you drive 100 km at each of the two speeds?

*Answer:*

a. The total distance driven = [(2 h )(40 km /h) + (2 h)( 60 km/h)] = 200 km

The total time = 2 + 2 = 4 h

average speed = (200 km)/(4 h) = 50km/h

b. total distance = 100 + 100 = 200 km

total time = [(100 km)/(40 km/h) + (100 km)/ 60 km/h)] = 4.17 h

average speed = (200 km)/(4.17 h) = 48 km/h

**(average velocity)** = displacement ÷ time.

**Velocity** is a vector. Both direction and quantity must be stated. It one train has a velocity of 100km/h north, and a second train has a velocity of 100km/h south, the two trains have different velocities, even though their speed is the same. Other examples of vectors are force, and field intensity.

If a person walked 400 m in a straight line in 5 min, that person's velocity would be (400 m [forward])÷(5 min) = 80 m/min [forward] .

If the same person walked 100 m [North] then 300 m [South] in 5 minutes, we first find their displacement.

displacement = 200 m [S]

velocity = 200÷5 = 40 m/min [S]

If that person walked 100 m [E] in .75 min, 100 m [N] in 1.50 min, 100 m [W] in 1.00 min and finally 100 m [S] in 1.75 min, that person would end up back where they started. Since their displacement is zero, their velocity is zero.

Remember,

**(average velocity) = displacement ÷ time.**

A hiker traveled 80.0 m [S] at 1.00 m/s, then 80.0 m [S] at 5.00 m/s. What is the hiker's average velocity?

*Answer:*

displacement = 160.0 m [S]

time for the first part is 80.0÷1.00= 80.0 s, time for the second part is 80.0 m ÷ 5.00 m/s = 16.0 s.

Total time = 80.0+16.0 = 96.0 s

Therefore, the velocity is (160.0 m [S])÷96.0 s = 1.67 m/s [S]

A train on a straight track traveled 60.0 km/h [E] for 2.00 h, stopped for 15 min, then traveled 100.0 km [W] at 133 km/h.

a. What was the train's average speed for the whole trip?

a. What was the train's average velocity for the whole trip?

*Answer:*

a. To find average speed, we need total distance and total time.

During the first part of the trip, the train covered 60.0x2.00 = 120 km in 2.00 h.

During the second part of the trip the train traveled 0.00 km in 0.25 h.

During the third part of the trip, the train traveled 100.0 km in 0.75 h.

In total the train traveled 220 km in 3 .00 h.

Average speed = (220 km)÷3.00 = 73.3 km/h

b. To find average velocity, we need displacement and total time.

During the first part of the trip, the train covered 60.0x2.00 = 120 km [E] in 2.00 h.

During the second part of the trip the train traveled 0.00 km in 0.25 h.

During the third part of the trip, the train traveled 100.0 km [W] in (100.0 km ÷ 133 km/h) = 0.75 h.

The train's displacement was (120-100) = 20 km [E] in 3 .00 h.

Average velocity = (20 km [E])÷3.00h = 6.7 km/h [E]

A runner covers one lap of a circular track 40.0 m in diameter in 62.5 s. For that lap, what were her average speed and average velocity?

*Answer:*

average speed = (total distance)/(total time) = (π*40.0)/(62.5) = 2.01m/s

average velocity = displacement/time = 0/62.5 = 0 m/s

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**acceleration = (change in velocity) ÷ time**.

**Acceleration** is a vector when it refers to the rate of change of velocity. Acceleration is scalar when it refers to rate of change of speed. A car slowing down to stop at a stop sign is accelerating because its speed is changing. We might refer to this type of acceleration as deceleration or negative acceleration. A car going at a constant speed around a curve is still accelerating because its direction is changing.

A pitcher delivers a fast ball with a velocity of 43 m/s to the south. The batter hits the ball and gives it a velocity of 51m/s to the north. What was the average acceleration of the ball during the 1.0ms when it was in contact with the bat?

*Answer:*

acceleration = (v_{f} - v_{i})/t = ( 51m/s to the north - 43 m/s to the south)/(1.0x10^{-3}s)

Letting south be positive and north negative yields

acceleration = ( -51m/s - 43 m/s )/(1.0x10^{-3}s) = -94000 m/s/s

acceleration = 94000 m/s/s to the north

(See bottom of page for answers.)

1. An automobile travels on a straight road for 40 km at 30 km/h. It then continues in the same direction for another 40km at 60km/h. a) What is the average velocity of the car during this 80 km trip? b) What is the average speed.

2. A truck on a straight road starts from rest and accelerates at 2.0 m/s^{2} until it reaches a speed of 20 m/s. Then the truck travels for 20 s at a constant speed until the brakes are applied, stopping the car in a uniform manner in an additional 5.0 s. How long is the truck in motion and what is its average velocity during the motion?

3. While driving home from school you travel at 95 km/h for 130 km then slow to 65 km/h. You get home in 3 hours and 20 min. How far is your hometown from school and what is the average speed?

4. A person takes a trip, driving with a constant speed of 89.5 km/h, except for a 22 min rest stop. If the person's average speed is 77.8 km/h, how much time is spent on the trip and how far does the person travel?

5. From t = 0 to t = 4.21 min, a man stands still, and from t = 4.21 min to t = 8.42 min, he walks briskly in a straight line at a constant speed of 1.91 m/s. In the time interval 1.00 min to 5.21 min what are

a. his average velocity

b. his average acceleration?

6. A bird watcher meanders through the woods, walking 0.684 km due east, 0.486 km due south, and 3.56 km in a direction 61.7 degrees north of west. The time required for this trip is 1.124 h. Determine the bird watcher's (a) displacement and (b) average velocity.

7. A truck driver is in a rush to pick up a load of eggs. She travels 40 miles at 80 mi/h, returning with a full truck along the same route at 40 mi/h. What was her average speed for the trip?

8. A race car driver sets out on a 100-mile race. At the halfway marker, her pit crew radios that she has averaged only 50 mi/hr. How fast must she drive over the remaining distance in order to average 100 mi/hr for the entire race?

9. In reaching her destination, a backpacker walks with an average velocity of 1.34 m/s, due west. This average velocity results because she hikes for 6.44 km with an average velocity of 2.68m/s, due west, turns around, and hikes with an average velocity of 0.447 m/s, due east. How far east did she walk?

10. You jog at 6 mi/h for 5 mi, then you drive another 5 mi in a car. With what average speed must you drive if your average speed for the entire 10 miles is to be 10.8 mi/hr?

11.

a. The distance of one lap around an oval track is 1.50 km. If a biker going at a constant speed makes one lap in 1.10 min, what is the speed of the bike and biker in meters per second?

b. Is the velocity of the bike also constant? Explain.

(See bottom of page for answers.)

1. If a bug's position is given by

x= 4m-(12m/s)t+(3m/s^{2})t^{2} (where t is in seconds and x is in meters.), what is its velocity at t = 1 s?

2. A UFO is traveling with a velocity of 3250 m/s. Suddenly the retro rocket is fired, the UFO slows to a stop with an acceleration whose magnitude is equal to 10m/s^{2}. What is the velocity of the UFO when the displacement of the craft is 215 km, relative to the point where the retro rocket began firing?

3. A vehicle traveling 60 km/h approaches an intersection just as the traffic light turns yellow. The yellow light lasts only 2.0 s before turning to red. The distance to the near side of the intersection is 30 m and the intersection is 15 m wide. The vehicle can decelerate at -6.4 m/s^{2} , whereas it can accelerate from 60 km/h to 70 km/h in 3.0 s.

a. If the brakes are applied, how far will the vehicle travel before stopping?

b. If the vehicle accelerates, how far will it travel before the light turns red?

4. A car accelerates at a rate of 0.6 m/s^{2}. How long does it take for this car to go from a speed of 55 mi/h to 60 mi/h?

5. A jalopy with an initial speed of 23.7 km/h accelerates at a uniform rate of 0.92 m/s^{2} for 3.6 s. Find the final speed and the displacement of the jalopy during this time.

6. A jet acquires a lift-off speed of 112 m/s in 20.0 s, starting from rest and traveling due east. What are the magnitude and direction of its average acceleration?

7. A bobsledder starting from rest accelerates uniformly down a 30 degree hill at 3.30 m/s^{2}. How long will it take to reach the bottom of the hill if her change in elevation is 110 m?

8. Two markers separated by 0.30 km placed on a road. A gas guzzler passed the first marker with a velocity of 5.0 m/s [E] and passed the second marker with a velocity of 33.0 m/s [E]. Calculate the car's average acceleration.

9. A jalopy is decelerated from 48m/s to 12m/s over 5 s. What is the displacement of the jalopy during this time?

10. How much time will it take a jalopy, starting from rest, to reach a velocity of 24 m/s over a distance of 315 m?

11. A jalopy travels on a long straight level road at 50 km/hr and then speeds up to 90 km/hr in 15 seconds. Calculate the jalopy's acceleration in m/s/s.

12. A kangaroo is moving with a velocity of 5.0 m/s at an angle of 30º North of East. The kangaroo, 4.0 seconds later, is moving with a velocity of 6.0 m/s at an angle of 45º South of East. What is the kangaroo's average acceleration?

13. A bobsledder is accelerating down a 30.0º hill at a = 2.80 m/s^{2}.

a. What is the vertical component of his acceleration?

b. How long will it take her to reach the bottom of the hill, assuming she starts from the rest and accelerates uniformly, if the elevation change is 315 m?

14.Suppose the position vector for a bug is given as a function of time by r(t)=x(t)+ y(t), with x(t)=at+b and

y(t)=ct^{2} + d, where a=1.00 m/s, b=1.00m, c=0.125 m/s^{2}, and d=1.00m.

Determine an equation for the instantaneous velocity as a function of t.

15. A thingamajig moves along the x axis according to equation x(t)=(2.60t^{2}-2.00t+3.00)m, where t is in seconds and x is in m.

a. Find the average speed between t = 1.90s and 2.90 s.

b. Find instantaneous speed at t = 1.90 s and 2.90 s.

c. Find average acceleration between t=1.90s and t = 2.90s.

d. Find instantaneous acceleration at t = 1.90s and 2.90s.

16. A driver traveling 40 km/h sees a child run onto the road 13 m ahead of his jalopy. He applies the brakes, and the jalopy decelerates uniformly at 8.0 m/s/s. The driver's reaction time is 0.25 s. Will the jalopy hit the child?

17. A jalopy traveling 70 km/hr comes to a stop at 120 m. Find the jalopy's acceleration.

(See bottom of page for answers.)

1. A biologist runs in a straight line toward his car at a speed of 4.0 m/s. The car is a distance d away. A hungry bear is 23 m behind the biologist and chases him at 6.0 m/s. The biologist reaches the car safely. What is the maximum possible value for d?

2. Two marbles are 400 cm apart. The blue marble starts traveling 26 cm/s toward the blue marble. The red marble starts traveling 4.0s later toward the blue marble at 31.25 m/s. Approximately how many seconds is the red marble traveling before they collide head-on? How far does each marble travel from their starting position?

3. A woman is running at a constant velocity of 5 m/s in an attempt to catch a stopped bus 11m ahead of her. However, it pulls away from her with a constant acceleration of 1m/s^{2}. How much time does it take her to reach the bus if she keeps running at the same velocity?

4. Two sprinters finish with times of 3:53.58 and 3:55.66. Assuming that both run 1609 m at constant velocity,, what distance separates them at the end of the race?

5. A woman on a bridge 86.8 m above a stream sees a bottle floating along at a constant speed. She drops a stone from rest when the bottle has 5.88 m more to travel before passing under the bridge. The stone hits the water 2.18 m in front of the bottle. Calculate the speed of the bottle.

6. After a refueling stop a race car accelerates at 6m/s^{2}, and after 4 s reenters the raceway. At that instant, another race car traveling at a constant speed of 70 m/s overtakes and passes the refueled car. If the refueled car maintains its acceleration, how much time is required for it to catch the other car?

7. When you are 20m away from a stopped hummer it begins accelerating away from you at 3 m/s/s. With what constant velocity should you run to catch the hummer?

8. A scooter traveling at 1.00 m/s passes a stopped cyclist. The instant the scooter passes the cyclist, the cyclist begins to accelerate at 0.200 m/s^{2} in the direction of the scooter.

a. How long does it take the cyclist to catch up to the scooter?

b. What is the speed of the cyclist when he catches up to the scooter?

c. How far do they travel before they meet?

(See bottom of page for answers.)

1. A clunker had an acceleration, a, and a deceleration of 1.84*a. The clunker must travel a short distance, L, in the minimum time. Starting and ending at rest, at what portion of L should the driver start braking?

2. A sprinter starting a 100-yard race accelerates from rest at 9 ft/s^{2}. After acquiring top speed, he runs at constant speed. If he finishes in 11s, how far does he run while accelerating?

3. An elevator travels upward as described by:

v(t)= (3.0 m/s^{2})t +(0.20m/s^{3})t^{2}

What is the acceleration of the elevator 4.0 s after starting from rest?

4. A 1350 kg four by four moving with a velocity of 15.0 m/s begins to accelerate at (1.5t + 2.60) m/s^{2}. Determine (a) its velocity and (b) its position after 6.00 seconds. (c) What resultant force acted on the four by four at t =3.00s?

5. A convoy travels at a speed of 72 km/h. The convoy must stop for 2.0 minutes. If the convoy slows down at a uniform rate of 1.0 m/s^{2} and, after the stop, speeds up at a rate of .50 m/s^{2}, how much longer is the time of the trip compared to the time if the convoy had kept a constant speed?

6. Solve m(d^{2}x/dt^{2})+b(dx/dt)-kx=0 given that it is an equation of motion of a body not in damped vibration. Translate the equation.

Selected solutions are printed below.

For solutions to all the problems on this page click here.

A10.

The following table summarizes what we know:

jog | drive | total/average | |
---|---|---|---|

d (mi) | 5 | 5 | 10 |

t (h) | t_{j} = ? |
t_{tot} - t_{j} = ? |
t |

v (mi/h) | 6 |
v |
10.8 |

Since 10 miles were to be covered at 10.8 mi/h, the time for the trip, t_{to}, is found by

t_{tot} = d _{tot} / v _{avg} = 10 / 10.8 = 0.926 h

Since 5 mi were jogged at 6 mi/h, the time spent jogging was

t_{j} = 5 / 6 = 0.833 h

The time left to drive the remaining 5 mi is therefore

0.926 - 0.833 = 0.0926 h

Since there are 5 mi to drive, the required average speed is

v_{d} = 5 / 0.0926 = 54 mi/h

For solutions to all the problems on this page click here.

B17.

(70 km/h)(1000 m/km)(1h / 3600s) = 19.4 m/s

Applying v_{f}^{2} = v_{i}^{2} + 2ad:

0^{2} = (19.4)^{2} + 2a(120)

a = -1.58 m/s/s

For solutions to all the problems on this page click here.

D5.

a. time lost during deceleration:

initial speed = 72 km/h*1000/3600) = 20 m/s

final speed = 0

t = (0-20)/(-1.0 m/s/s) = 20 s

distance traveled = v_{i}t +(1/2)at^{2} = (20 m/s)(20 s) +(1/2)(-1.0)(20)^{2} = 200 m

time to travel 200 m if the train maintained 20 m/s = 200/20 = 10 s

time lost = 20 - 10 = 10 s

b. time lost during stop = 2.0 minutes = 120 s

c. time lost during acceleration

initial speed =0

final speed = 20 m/s

t = (20 -0)/(0.50 m/s/s) = 40 s

distance traveled = v_{i}t +(1/2)at^{2} = (0 m/s)(40 s) +(1/2)(0.50)(40)^{2} = 400 m

time to travel 400 m if the train maintained 20 m/s = 400/20 = 20 s

time lost = 40 - 20 = 20 s

d. total time lost

10 + 120 + 20 = 150 s or 2.5 minutes

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