Physics
Force and Motion Test
Question Explanations
Objective
• To review the questions contained in your test to understand how to
solve any questions that you may have missed
• The question number listed here are arbitrary and do not reflect any
particular test (questions were randomized)
Question 1
• Usain Bolt ran the 200 m in 19.19 s. What is the average speed of his
run?
• All we need for this problems is v = d/t
• 200 m/ 19.19s = 10.42 m/s
Question 2
• A bird's egg falls from a nest that is 18 m above the ground. How long
will it take that egg to hit the ground (Neglect air resistance)?
• We know that y= 18m
• We also know that a = 9.81m/s2 AND that vi = 0 the instant it starts to
fall
• We are trying to find time
• Let’s see: y, a, vi, and t
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•
•
•
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y = vit + ½ at2
18 = 0 + ½ (9.81)t2
18 = 4.905t2
18/4.905 = t2
t = 1.92 s
Question 3
• A boy is trying to throw a rose to a girl he likes who lives on the second
floor (about 8m high). How fast will he need to throw the flower straight
up so that its highest point is the same as the height of the second floor?
• We know that y = 8
• We know that vf = 0 since it stops momentarily at the top of its height
• We know that a = -9.81 m/s2 (since gravity is in the opposite direction of
travel
• We need to find vi
• Let’s see: y, vf, a, vi
• vf2 = vi2 + 2ay
Question 3, continued
• 0 = vi2 + 2(-9.81)(8)
• 0 = vi2 + -156.96
• 156.96 = vi2
• vi = 12.5 m/s
Question 4
• A car starts from rest and accelerates at a constant rate in a straight
line. In the first second, the car covers a distance of 2.0 meters. How
much additional distance will the car cover during the second second
of its motion?
• For the first second, we know
• x = 2 m, vi = 0 m/s, t = 1 s
• It says that the car is accelerating but not by how much. Let’s find out
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•
•
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x = vit + ½ at2
2 = ½ a (1)2
2=½a
a = 4 m/s2
Question 4, continued
• Since it is accelerating, it is going faster and faster
• It DOES NOT cover the same distance in each second
• How far does it travel from start to the 2s mark?
• x = ?, a = 4 m/s2, vi = 0 m/s, t = 2s
• x = vit + ½ at2
• x = 0 + ½ (4) (2)2
• x=8m
• Since it would have covered 2 m in the first second, then it would have
covered 6 m in the second second.
Question 5
• Carl Lewis set a world record for the 100.0 m run with a time of 9.86 s.
If, after reaching the finish line, Mr. Lewis walked directly back to his
starting point in 90.9 s, what is the magnitude of his average velocity
for the 200.0 m?
• This checks for understanding of concepts
• If he went 100m and then came back 100m, then his displacement is
zero
• Since velocity is displacement divided by time, then average velocity
was zero
Question 6
• What is the velocity of the object at t = 7.0 s?
• The velocity is equal to the slope of the line at that time
• m = rise/run  (0-40)/(6-5)  -40/1 = -40 m/s
Question 7
• At a drag race, a car accelerates for 2.6 s to a speed of 112 m/s.
Maintaining that top speed, it reaches the finish line 400 m from the
starting line. Once there, a parachute deploys enacting an
acceleration of 39 m/s2.
• A) What is the acceleration of the car when the race starts?
• B) How much time does it take the car to reach the finish line?
• C) How much time does it take to stop?
• D) How far from the starting line does the car finally stop?
Question 7 part A
• We know that: t= 2.6 s, vf = 112 m/s
• We also know that vi = 0 m/s
• We can use the formula a = (vf – vi) / t
• a = (112 – 0)/ 2.6
• a = 43 m/s2
Question 7 part B
• It accelerates in the first 2.6 s
• It doesn’t stop there, it keeps travelling forward BUT NOT accelerating
after 2.6s
• How far did it travel in the first 2.6s?
• x = vit + ½ at2
• x = 0 + ½ (43)(2.6)2
• x = 145 m
• The finish line is 400 m from the starting line
• 400 – 145 = 255 m left to go
• It is travelling at a constant speed (112 m/s)
• v = d/t  t = d/v  255 / 112 = 2.3 s
• The entire trip would have taken 2.6 + 2.3 = 4.9s
Question 7 part C
• Once it crosses the line, the parachute slows it down
• For that portion of the trip, we know:
• a = -39 m/s2 (because it is slowing down), vi = 112 m/s, and vf = 0 m/s
• We are trying to find time
• Once again, we can use: a = (vf – vi) / t
• -39 = (0 – 112)/t
• -39 t = -112
• t = 2.9s
Question 7 part D
• We already know that it travelled 400m to get to the finish line
• Now we need to know how much farther it travelled
• Once again, we know that: a = -39 m/s2, vi = 112 m/s, and vf = 0 m/s
• We need to find x
• vf2 = vi2 + 2ax
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•
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0 = 1122 + 2(-39)x
0 = 12544 -78x
-12544 = -78 x
x = 161 m
• Total distance 400 + 161 = 561 m
Question 8
• A runaway train is barreling down the tracks at a speed of 75 km/h compared to
the tracks. A solution has been devised to send another train down a parallel
track to catch up and stop the train. The second locomotive will not pull cargo
and can achieve a speed of 90 km/h compared to the track. The first train has a 5
km head start. Once the train catches up, it will take 15 minutes to slow the train
to a stop
• A) How fast would the second train travel compared to the first?
• B) How long will it take the second train to catch up to the second in minutes?
• C) What would the deceleration of first train be to slow to a stop in 15 minutes.
• D) What is the total distance that the first train will travel after being caught up
to by the second?
Question 8 part A
• Compared to each other, one train is travelling 15 km/h
Question 8 part B
• We know that there is a 5 km head start
• Since the relative speed of the 2nd train is 15 km/h compared to the first,
you can work the problem as if the first train were not moving
• v = d/t  t = d/v  5/15 = 1/3 of an hour
• The question asks for minutes: 1 hour/3 times 60 minutes/hour = 20
minutes
Question 8 part C
• When the “target” train starts to slow down, we know:
• vi = 75 km/h, t = 15 minutes, vf = 0 km/h
• Since, vi is in km/h and time is in minutes, you will have to convert one or
the other (or both) so that they have the same units
• NOTE: for your answer key, I made 3 different versions of this same answer
• One method would be to change time to hours: t=.25 hours
• We can use: a = (vf – vi) / t
• a = (0 – 75)/.25 = - 300 km/h2
Question 8 part D
• What we know
• vi = 75 km/h; a = -300 km/h2, vf = 0 km/h
• We are trying to find x
• vf2 = vi2 + 2ax
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•
•
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0 = 752 + 2(-300)x
0 = 5625 -600x
-5625 = -300 x
x = 9.4 km
Question part D, alternate
• You may have decided to use time instead
• vi = 75 km/h, a = - 300 km/h2, t = .25 h
• x = vit + ½ at2
• x = (75)(.25) + ½ (-300)(.25)2
• x = 18.75 – 9.375
• x = 9.4 km