Physics
Name: ______________________
Solutions to Problem Solving Practice
Constant Speed With No Conversions:
4. What must your average speed be to travel 5.0 m in 2.0 s?
a. Make a list of the known and unknown.
x = 5m
t = 2.0s
v =?
b. Using the symbols we have introduced, write down the equation you will use to find the average speed:
€
s=
dtot
t tot
c. Rewrite the equation with numbers substituted for all the quantities whose values are known:
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s=
5.0m
2.0s
d. Calculate the car’s average speed, and put the complete (rounded, labeled) answer in a box.
s = 2.5m /s
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5. During a sneeze, your eyes shut for 0.50 s. If you are driving a car at 24.0 m/s, how far does the car
move during your sneeze?
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List of known and unknown:
Δt = 0.5s
v = 24.0m /s
Equation in symbol form solved for the unknown:
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Substitution and solve:
Δx = ?
v=
Δx
becomes : Δx = v Δt
Δt
Δx = 24.0m /s × 0.5s
Δx = 12m
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€
Constant€Speed With Conversions:
6. How long does it take a snail to move 3.5 m if his average speed is 1.00 cm/minute?
Δx = 3.5m
v=1cm/min =1.6667x10-4m/s
Δt =?
v = Δx/Δt
Δt = Δx/v
Δt = 3.5m/1.6667x10-4m/s
Δt = 20999.58
Δt = 2.1x104s
7. What must your average speed be (in m/s) if you travel 30.25 km in 30.0 minutes?
Δx = 30.25km = 30,250m
Δt = 30.0min = 1800s
v=?
v = Δx/Δt
v = 30,250m/1800s
v = 16.8056m/s
v = 16.8m/s
Shifting Frame of Reference:
9. (Problem 12 pg. 39) A car traveling 88 km/h is 110 m behind a truck traveling at 75 km/h. How
long will it take the car to reach the truck?
Δx = 110m
v = 13km/h = 3.6111m/s
t=?
t = Δx/v
t = 110m / 3.61111m/s
t = 30.4615s
t = 30s
Average Speed Problems:
10. Example: A car travels 200 km for 1.0 h, then it travels at a speed of 85 km/h for 1.5 h, and then it
travels a distance of 64 km at a speed of 120 km/h. What was its average speed?
Solution: You must find the total distance and total time by adding up the three distances and times for the three legs
of the journey. For leg 1, they are given. But you must solve for distance for leg 2 and time for leg 3 before you can complete
the problem.
Leg 1:
∆x = 200km = 200,000m
∆t = 1h = 3,600s
v = 55.5556 m/s
Leg 2:
∆x = ?
∆x = v∆t
∆t = 1.5h = 5,400s
Δx = (23.6111m/s)(5,400s) = 127,500m
v = 85km/h = 23.6111m/s
Δx = 1.275x105m
Leg 3:
∆x = 64,000m
∆t = ∆x/v
∆t = ?
Δt = 64,000m / 33.33m/s
v = 120km/h = 33.33m/s
Δt = 1920.1920s
Final step:
average speed =
total distance
=
total time
200,000m + 127,500m + 64,000m
= 36m/s
3,600s + 5,400s + 1,920s
11. Anne in C Building is walking to class. She walks 3 m/s for 10 s, then stops to talk to Dave for
5s. Realizing that she is late to class, she sprints at 6 m/s for 4 s to reach her class. What was her
average speed for the entire trip?
Part 1:
Δx = ?
Δt = 10s
Δx = vΔt
Δx = (3m/s)(10s)
Δx = 30m
Part2:
Δx = 0m
Δt = 5s
Part 3:
Δx = ?
Δt = 4s
Δx = vΔt
Δx = (6m/s)(4s)
Δx = 24m
v = 3m/s
Δv = 0m/s
v = 6m/s
d
30m + 0m + 24m 54m
average speed = tot =
=
= 2.8m / s = 3m/s
t tot
10s + 5s + 4s
19s
Average Velocity Problem:
12. A runner dashes from the starting line (x = 0) to a point 84 m away and then turns around and runs to a
point 18 m away from the starting point in 23 seconds. What is the average velocity?
x 0 = 0m
x f = 18m Δt = 23s
v =?
Δx 18m
v=
=
Δt 23s
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v = 0.78m /s away from starting line
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Average
€ Velocity Problem with too much info given (read carefully):
13. Kelly is standing at the open window of her physics room, 20m above the ground. She throws a ball
straight up at an initial speed of 14 m/s, it rises another 10m to a total height of 30m above the ground in
1.43s. It then falls to the ground reaching a speed 24.2 m/s before it hits in another 2.47s. Calculate the
average velocity of the ball.
x f = 0m
Δx
Equation in symbol form: v =
Δt
List:
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x 0 = 20m
Δt =1.43s +2.47s = 3.9s
Substitution and solve (since velocity is a vector you need to include direction in your final answer):
v=
x f − x i 0m − 20m
=
= [5.1m /s down]
t f − t i € 3.9s − 0s
Average Acceleration Problem:
€
14. A car that is traveling west at 15.0 m/s accelerates for 3.0 seconds until it reaches a velocity of 35 m/s.
List:
€
v o = 15m /s
v f = 35m /s
Δt = 3s
Δv v f − v 0 35m /s2 −15m /s2
Equation in symbol form: a =
=
=
Δt t f − t 0
3s
2
a = 6.7m /s west
15. Calculate the acceleration of Kelly’s ball in question 13 for the trip up to the peak and then calculate
€ of the ball as it falls to the ground. If done correctly, you will get -9.8m/s2 for both. Be
the acceleration
€
very careful with the signs of your velocity! Remember that up is positive and down in negative.
Up
|
Down
|
v 0 = 14m /s v f = 0m /s
Δt = 1.43s
v 0 = 0m /s v f = −24.2m /s
Δt = 2.47s
a=
Δv v f − v 0 0 −14m /s
=
=
=
Δt t f − t 0
1.43s
[a = −9.8m /s
2
a=
|
= 9.8m /s2 down ]
[a = −9.8m /s
|
|
€
Δv v f − v 0 −24.2 − 0m /s
=
=
=
Δt t f − t 0
2.47s
€
2
= 9.8m /s2 down ]