Problem#1
A car travels in the +x-direction on a straight and level road. For the first 4.00 s of its motion, the average velocity of the car is vav-x = 6.25 m/s. How far does the car travel in 4.00 s?
Answer:
We know the average velocity is 6.25 m/s.
Then, Δx = vavgΔt = 25.0 m
Problem#2
In an experiment, a shearwater (a seabird) was taken from its nest, flown 5150 km away, and released. The bird found its way back to its nest 13.5 days after release. If we place the origin in the nest and extend the +x-axis to the release point, what was the bird’s average velocity in m/s. (a) for the return flight, and (b) for the whole episode, from leaving the nest to returning?
Answer:
13.5 days = 1.166 × 106 s. At the release point, x = +5.150 × 106 m.
(a) the average velocity in m/s for the return flight is
vavg = ∆x/∆t = (0 – 5.150 × 106 m)/1.166 × 106 s = –4.42 m/s
(b) For the round trip, x2 = x1 and Δx = 0. The average velocity is zero.
The average velocity for the trip from the nest to the release point is positive.
Problem#3
Trip Home. You normally drive on the freeway between San Diego and Los Angeles at an average speed of 105 km/h (65 mi/h) and the trip takes 2 h and 20 min. On a Friday afternoon, however, heavy traffic slows you down and you drive the same distance at an average speed of only 70 km/h (43 mi/h). How much longer does the trip take?
Answer:
Target variable is the time Δt it takes to make the trip in heavy traffic.
Use the information given for normal driving conditions to calculate the distance between the two cities
vavg = ∆x/∆t
∆x = vavgt = (105 km/h)(1 h/60 min)(140 min) = 245 km
Now use vavg-x for heavy traffic to calculate Δt; Δx is the same as before:
∆t = ∆x/vavg = 245 km/(70 km/h) = 3.50 h = 3 h and 30 min.
The trip takes an additional 1 hour and 10 minutes.
Problem#4
From Pillar to Post. Starting from a pillar, you run 200 m east (the +x-direction) at an average speed of 5.0 m/s and then run 280 m west at an average speed of 4.0 m/s to a post. Calculate (a) your average speed from pillar to post and (b) your average velocity from pillar to post.
Answer:
The post is 80 m west of the pillar. The total distance traveled is 200 m+ 280 m = 480 m.
(a) The eastward run takes time 200 m/(5.0 m/s) = 40 s and the westward run takes 280 m/(4.0 m/s) = 70 s. The average speed for the entire trip is
Savg = 480 m/110 s = 4.4 m/s
(b) your average velocity from pillar to post is
vavg = ∆x/∆t = (200 m – 280 m)/110 s = –0.73 m/s, the average velocity is directed westward.
Problem#5
Starting from the front door of your ranch house, you walk 60.0 m due east to your windmill, and then you turn around and slowly walk 40.0 m west to a bench where you sit and watch the sunrise. It takes you 28.0 s to walk from your house to the windmill and then 36.0 s to walk from the windmill to the bench. For the entire trip from your front door to the bench, what are (a) your average velocity and (b) your average speed?
Answer:
(a) Let +x be east. Δx = 60.0 m− 40.0 m = 20.0 m and Δt = 28.0 s + 36.0 s = 64.0 s. So, the average velocity is
vavg = ∆x/∆t = 20.0 m/64.0 s = 0.312 m/s
(b) the average speed is
Savg = (60.0 m + 40.0 m)/64.0 s = 1.56 m/s
The average speed is much greater than the average velocity because the total distance walked is much greater than the magnitude of the displacement vector.
Problem#6
Honda Civic travels in a straight line along a road. Its distance x from a stop sign is given as a function of time t by the equation x(t) = αt2 – βt3, where α = 1.50 m/s2 and β = 0.0500 m/s3 . Calculate the average velocity of the car for each time interval: (a) t = 0 to t = 2.00 s, (b) t = 0 to t = 4.00 s and (c) t = 2.00 s to t = 4.00 s?
Answer:
the equation x(t) = αt2 – βt3 = (1.50 m/s2)t2 – (0.0500 m/s3)t3, then x(0) = 0, x(2.00 s) = 5.60 m, and x(4.00 s) = 20.8 m
(a) the average velocity of the car for each time interval t = 0 to t = 2.00 s is
vavg = ∆x/∆t = (5.60 m – 0)/2.00 s = 2.80 m/s
(b) the average velocity of the car for each time interval t = 0 to t = 4.00 s is
vavg = ∆x/∆t = (20.80 m – 0)/4.00 s = 5.20 m/s
(c) the average velocity of the car for each time interval t = 2.00 s to t = 4.00 s is
vavg = ∆x/∆t = (20.80 m – 5.60 m)/2.00 s = 7.60 m/s
The average velocity depends on the time interval being considered.
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