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Physics Help >> Index to Physics Homework Help » Speed, Velocity, Acceleration
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.
Example 2: 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
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.
Example 4: 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?
Example 6: 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
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.
Example 7: 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 = (vf - vi)/t = ( 51m/s to the north - 43 m/s to the south)/(1.0x10-3s)
Letting south be positive and north negative yields
acceleration = ( -51m/s - 43 m/s )/(1.0x10-3s) = -94000 m/s/s
acceleration = 94000 m/s/s to the north
PROBLEMS
A. Average Velocity, Average Speed
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/s2 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 woman makes a trip to a nearby shopping mall that is located 40 miles from her home. On the trip to the mall she averages 80 mi/hr but gets a speeding ticket upon her arrival. On the return trip she averages just 40 mi/hr. What was her average speed for the entire 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 dirt-bike track is 1.50 km. If a rider going at a constant speed makes one lap in 1.10 min, what is the speed of the bike and rider in meters per second? b. Is the velocity of the bike also constant? Explain.
B. Constant Acceleration
1. If a particle's position is given by
x= 4m-(12m/s)t+(3m/s2)t2 (where t is in seconds and x is in meters.)
What is its velocity at t = 1 s?
2. A spacecraft is traveling with a velocity of 3250 m/s. Suddenly the retrorocket is fired, and the spacecraft begins to slow down with an acceleration whose magnitude is equal to 10m/s2. What is the velocity of the spacecraft when the displacement of the craft is 215 km, relative to the point where the retrorocket began firing?
3. A person driving her car at 60 km/h approaches an intersection just as the traffic light turns yellow. She knows that the yellow light lasts only 2.0 s before turning to red, and she is 30 m away from the near side of the intersection. The intersection is 15 m wide. Her car's maximum deceleration is -6.4 m/s2 , whereas it can accelerate from 60 km/h to 70 km/h in 3.0 s. Ignore the length of her car and her reaction time. a) If she hits the brakes, how far will she travel before stopping?
b) If she hits the gas instead, how far will she travel before the light turns red?
4. A certain car is capable of accelerating at a rate of 0.6 m/s2. How long does it take for this car to go from a speed of 55 mi/h to a speed of 60 mi/h? 5. A car with an initial speed of 23.7 km/h accelerates at a uniform rate of 0.92 m/s2 for 3.6 s. Find the final speed and the displacement of the car during this time.
6. The Concorde jetliner achieves 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 skier is accelerating down a 30 degree hill at 3.30 m/s2. How long (in sec.) will it take her to reach the bottom of the hill if the elevation change is 110 m? She starts from rest and accelerates uniformly.
8. In an acceleration test for a sports car, two markers 0.30 km apart were set up along a road. The car 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 between the markers.
9. A car is decelerated from 48m/s to 12m/s over 5 s. What is the displacement of the car during this time?
10. How much time will it take a car, starting from rest, to reach a velocity of 24 m/s over a distance of 315 m?
11. A car travels on a long straight level road at 50 km/hr and then speeds up to 90 km/hr in 15 seconds to pass a slower-moving vehicle. Calculate the car's acceleration in m/s/s.
12. A student is moving with a velocity of 5.0 m/s at an angle of 30º North of East. The student is moving 4.0 seconds later with a velocity of 6.0 m/s at an angle of 45º South of East. what is the student's average acceleration?
13. A particle moving along the x-axis has its velocity described as v = 2t2 m/s. Its initial position at t = 0 is x0 = 1 m. At t = 2 s, what is the product of the particle's position with its acceleration?
14.Suppose the position vector for a particle is given as a function of time by r(t)=x(t)+ y(t), with x(t)=at+b and
y(t)=ct2 + d, where a=1.00 m/s, b=1.00m, c=0.125 m/s2, and d=1.00m. Determine an equation for the
instantaneous velocity as a function of t.
15. An object moves along the x axis according to equation x(t)=(2.60t2-2.00t+3.00)m, where t is in seconds.
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. The speed limit in a school zone is 40 km/h. A driver traveling at this speed sees a child run onto the road 13 m ahead of his car. He applies the brakes, and the car decelerates at a uniform rate of 8.0 m/s/s. If the drivers reaction time is 0.25 s, will the car stop before hitting the child?
17. A car traveling 70 km/hr comes to a stop at 120 m. Find the car's acceleration.
18.A Skier is accelerating down a 30.0º hill at a = 2.80 m/s2. 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?
C. Two Objects
1. A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.0 m/s. The car is a distance d away. The bear is 23 m behind the tourist and running at 6.0 m/s. The tourist reaches the car safely. What is the maximum possible value for d? 2. Two toy cars are 400 cm apart and begin to move towards each other. The first car requires 3.2 seconds to travel 100 cm, and the second car can move 130 cm in 5 seconds. The slower car gets a 4.0 second head start (starting from the same position) How much time is needed for the faster car to catch the slower car? How far does each travel this time? 3. A bus has stopped to pick up riders. A woman is running at a constant velocity of 5 m/s in an attempt to catch the bus. However when she is 11 m from the bus, it pulls away with a constant acceleration of 1m/s2. From this point, how much time does it take her to reach the bus if she keeps running at the same velocity?
4. Two runners in a 1609 m race finish with times of 3:53.58 and 3:55.66. Assuming that both run at their average speeds during the entire time, what distance separates them at the end of the race? 5. A woman on a bridge 86.8 m high sees a raft floating at a constant speed on the river below. She drops a stone from rest in an attempt to hit the raft. The stone is released when the raft has 5.88 m more to travel before passing under the bridge. The stone hits the water 2.18 m in front of the raft. Calculate the speed of the raft.
6. A race driver has made a pit stop to refuel. After refueling he leaves the pit area with an acceleration whose magnitude is 6m/s2, and after 4 s he enters the main speedway. At the same instant, another race car that is on the speedway and traveling at a constant speed of 70 m/s overtakes and passes the entering car. If the entering car maintains its acceleration, how much time is required for it to catch the other car?
7. When you are 20m away from your bus it begins accelerating at 3 m/s/s (from rest). With what constant velocity should you run at to catch the bus?
8. A jogger traveling at 1.00 m/s passes a stopped cyclist. The instant the jogger passes the cyclist, the cyclist begins to accelerate at 0.200 m/s2 in the direction of the jogger.
D. Other
1. A car had a maximum acceleration, a, which remains constant to high speeds, and it gas a maximum deceleration of 1.84*a. The car must travel a short distance, L, starting and ending at rest, in the minimum time T. At what fraction of L should the driver move her foot from the gas pedal to the brakes?
2. In the one hundred yard dash, a sprinter accelerates from rest to top speed with an acceleration whose magnitude is 9 ft/s2. After achieving this speed, he runs the remainder of the race without speeding up or slowing down. if the total race is run in 11s, how far does he run in this acceleration phase.
3. An elevator starts from rest and travels upward with a speed that varies with time according to:
What is the acceleration of the elevator at t = 4.0 s?
4. A 1350 kg car is initially at the origin of coordinates and moving with a velocity of 15.0 m/s (i). The car then accelerates at (1.5t + 2.60)m/s2 (i). Determine (a) its velocity and (b) its position after 6.00 seconds. (c) What resultant force acted on the car at t =3.00s? 5. A train normally travels at a uniform speed of 72 km/h on a long stretch of straight, level track. on a particular day, the train must make a 2.0 minute stop at a station along this track. if the train decelerates at a uniform rate of 1.0 m/s2 and, after the stop, accelerates at a rate of .50 m/s2, how much time is lost because of stopping at the station?
6. Find the general solutions of the equation m(d2x/dt2)+b(dx/dt)-kx=0 noting that it is the equation of motion of a particle(not in damped vibration). Give the physical interpretation.
Answers
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A10. The following table summarizes what we know:
Since 10 miles were to be covered at 10.8 mi/h, the time for the trip, tto, is found by ttot = d tot / v avg = 10 / 10.8 = 0.926 h Since 5 mi were jogged jogged at 6 mi/h, the time spent jogging was tj = 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 vd = 5 / 0.0926 = 54 mi/h
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B17. (70 km/h)(1000 m/km)(1h / 3600s) = 19.4 m/s
Applying vf2 = vi2 + 2ad:
02 = (19.4)2 + 2a(120)
a = -1.58 m/s/s
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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 = vit +(1/2)at2 = (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 = vit +(1/2)at2 = (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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If the same person walked 100 m [North] then 300 m [South] in 5 minutes, we first find their displacement.