Grade 10 Exam Review #2- Motion Solutions
1. What are the two large categories of motion that encompass all possibilities?
Acceleration and Constant Velocity.
2. Identify the following as either scalar (S) or vector (V).
S-timeS-speed S-weight V-velocity
V-positionS-temperatureS-mass S-distance
V-force V-displacement V-acceleration
3. Define: position, distance, displacement, speed, velocity, acceleration
Position- Where something is located with reference to an origin
Distance- path taken without regard to direction.
Displacement- A change in position, that has a direction.
Speed- The rate of change of distance and time
Velocity- The rate of change of displacement and time.
Acceleration- The rate which velocity changes and time.
4. Describe the way in which science handles uncertainty in measurement. (sig figs and rules)
There is uncertainty in every measurement. In science, we have to have a way to handle this uncertainty in calculations to know how to report the proper accuracy and precision for every measurement. Scientists have agreed to use a set of rules to determine exactly how this works:
- ALL non-zero digits (1,2,3,4,5,6,7,8,9) are ALWAYS significant.
- ALL zeroes between non-zero numbers are ALWAYS significant.
- ALL zeroes which are SIMULTANEOUSLY to the right of the decimal point AND at the end of the number are ALWAYS significant.
- Trailing zeroes are not significant.
5. Write the number of sigfigs for the following:
_2_ a. 25 cg (1) _1_ e. 1000 kg (1)_3_ i. 1110 L (1)
_1_ b. 0.004 km (1,4)_3_ f. 1.00 m (1,3)_2_ j. 1.0 x 104km (1)
_3_ c. 3.20 X 10-3 mL (1,3) _3_ g. 1010 dg (1,2,4)_1_ k. 0.00001 m (1)
_4_ d. 10.07 kJ (1,2,4)_4_ h. 0.01010 km (1,2,3)_5_ l. 100.00 kg (1,3,4)
6. What are some types of numbers that do not obey the same rules for sigfigs? (Exceptions)
numbers in mathematical formulas, counted values, given values
7. Provide the answers to the correct decimal places/sigfigs for the following:
a. 5.9 m X 12.0 m = 71m(Area of a lawn)
b. 895 kg ÷ 15 students = 6.0 x 101kg per student(Average mass of student)
c. 254.62 g + 113.7 g + 147 g = 515g(Total mass of copper)
d. 38.5390 cm – 25.75 cm = 12.79cm(Length of tape)
8. Why is it so important in dimensional analysis to multiply by one? (ie)
Mathematically, dimensional analysis is an equation with a left side and right side. Without touching the left hand side of the equation, the only way to preserve the equality of both sides is to multiply by one which all the conversion factors are equivalent to.
9a) b) c)
d) e)
f)g) h)
10. A man walks 25 km in 5.0 hours. What is his average speed?
25km / 5hr = 5km/hr.
11. A train can travel at 150 km/hr. How long would it take to travel 600 km?
150km/hr x 4 = 600km
12. If a bird flies at 25 km/hr for 1.3 hours, how far did it travel?
25km x 1.3hr = 32.5km.
13. A man runs a 40.0 km marathon in a time of 2:20:37. Find his average speed in m/s.
40,000m / 4837s = 8.3m/s.
14. If a woman walks 2.5 km in 2 hours, how far could she walk in 3 hours if she walked twice as
fast?2.5km/2h = 1.25km/h1.25 km/h x 2 = 2.5 km/h 2.5km/h x 3 = 7.5km.
15. A car travels for 2.5 hours at 85.0 km/hr then 3.5 hours for 105.0 km/hr. What is
the car’s average speed?
16. A rocket travels at 25 000 km/hr upon reentry into the Earth’s atmosphere. How
far would it travel in 60.0 seconds?
so
17. If the speed limit is 110 km/hr and you have to travel 39.5 m, how long would it
take?
18. A caterpillar crawls 1.05 cm in 3.1 seconds. Find this speed in m/s and km/hr.
19. Explain the difference between speed and velocity.
Speed is the rate of distance over time ignoring direction, Velocity is the rate of displacement over time with regards to direction.
20. A car travels 20.0 km north and then 30.0 km south. What is the distance for the whole
trip? displacement? If the total time is 0.5 hours, what was the average speed? Average
velocity?
Distance= 50km. Displacement= 10km[S] Avg speed= 50 km/0.5 h = 100km/hr,
Velocity = 10km[S]/0.5h = 20km/h [S]
21. Describe the motion of the following graphs with respect to a position time graph.
a) at rest (positive position)b) constant positive velocity
c) constant negative velocityd) slowing down in a positive direction
e) speeding up in a positive directionf) slowing down in a negative direction
g) speeding up in a negative direction
22. Describe the motion of the following graphs with respect to a velocity time graph.
a) constant positive velocity b) constant positive acceleration
c) constant negative velocityd-g) non uniform acceleration
P/T Graph
a. What is the position at 1 s, 2s, 5s, 6.5s, 8s, 9.5s, 10s?
2m[N], 3m[N], 2m[N], 0m, 2m[S], 0m, 4m[N]
b. What is the displacement from 0-3 s? 3-6s? 6-9.5s? 0-9s?
6m[N], 4m[S], 0m, 2m[S]
c. What is the distance traveled from 0-3 s? 3-6s? 6-9.5s? 0-9s?
6m, 4m, 8m, 22m
d. What is the average velocity from 6-7 s? 9-10s? 5-10s? 2-8s?
4m/s[S], 8m/s[S], 0.4m/s[N], 1.2m/s[S]
e. What is the instantaneous velocity at 0.5s? 2s? 5.25s? 8s? 9.2s?
leave out
f. What is the average speed from 6-7 s? 9-10s? 5-10s? 2-8s?
4m/s, 8m/s, 0.4m/s, 1.2m/s
g. What is the instantaneous speed at 0.5s? 2s? 5.25s? 8s? 9.2s?
Leave out.
h. What time interval(s) is the object not moving?
7-9 s
i. What time interval(s) is the object accelerating?
1-6 s
j. What time interval(s) is the object moving at a constant velocity?
1-2s,6-10 s
24. Answer the questions for the following V/T graph.
i. What is the velocity at: (Need direction please)
a) 0 s: 10m/s [S]b) 10 s: 10m/s[S] c) 15 s: 20m/s [S] d) 20 s: 0m/s
e) 25 s: 30m/s [N]f) 30 s: 25m/s[N]
ii. What is the acceleration for the time interval: (Need direction)
a) 0-10 s: 0b) 10-15 s: 2m/s2 [S]c) 15-20 s:4m/s2 [N]
d) 20-25 s: 6m/s2 [N]e) 25-35 s: 1m/s2 [S]
iii. What is the displacement for the given time interval: (Need direction)
a) 0-10 s: 100m[S]b) 10-20 s: 125m [S]c) 20 – 25 s: 75m [N]
d) 0-20 s: 225m[S]e) 10-25 s: 50m[S]f) 0-30 s: 12.5m[S]
25. An object can accelerate at -4 m/s2. How much time from 32 m/s to 12 m/s?
t=(12-32)/-4t=5 seconds.
26. A motorcycle speeds up from rest to 11.0 m/s in 2.0 s. What is the average
acceleration of the motorcycle? a=(11-0)/2 a= 5.5m/s2
25. James accelerates at an average 2.5 m/s2 [N] for 3.5 s. What is his change in
velocity at the end of 1.5 s?Δv=2.5(3.5) Δv=8.8 m/s
26. A skateboarder rolls down a hill and changes his speed from rest to 3.9 m/s. If the
average acceleration down the hill is 0.45 m/s2, for how long was the skateboarder
on the hill? t=(3.9-0)/.45t= 8.7 s
27. Jo is moving at 1.8 m/s near the top of a hill, 2.2 s later she is travelling at 6.3
m/s. What is her average acceleration? a=(6.3-1.8)/2.2 a=4.1m/s.
28. A bus with an initial speed of 12 m/s accelerates at 0.62 m/s2 for 12 s. What is
the final speed of the bus? Vf=12+.62(12) Vf = 19.44m/s
29. A snowmobile reaches a final velocity of 22.5 m/s after accelerating at 2.2 m/s2
for 15 s. What was the initial speed of the snowmobile? Vi=22.5-2.2(15) = -10.5 m/s
30. An object is accelerating at 2 m/s2 from rest for 5.0 s. Find the distance travelled.
25m. (This was a tough one not to worry for exam!)
31. Trace the history of motion through Aristotle, Galileo and Newton.
Aristotle
Objects do not move without a force.
Objects in motion always require a force to keep them moving.
Objects seek their natural state, which is at rest.
Galileo
It is just as natural for a body to be in motion at a constant velocity than for it to be at rest
Without friction, an object would go on forever if given an initial force.
Newton
Came up with the term Inertia.
Objects do not change motion without unbalanced force.
Objects in motion do not always require a force to keep them moving.
Objects have two “natural” states of motion, at rest and moving at a constant speed and direction.
- State Newton’s three laws and give an example to illustrate each.
a.a.Objects will remain at rest or in uniform motion in a straight line unless acted upon by an unbalance force. (Bus stopping- When a bus stops and you are standing in it your inertia remains at rest {standing} so the bus moves from under your feet until you hit an unbalanced force {floor}.
a.b.Acceleration is directly proportional to force and inversely proportional to mass. (Kicking an object- Kicking a ball requires less force because it has less mass, while kicking a brick requires more force because it has more mass)
a.c.For every action there is an equal and opposite reaction. (Person walking- Foot pushes ground back and ground pushes foot forward)
33. Explain how a person can walk using Newton’s third law.
Foot pushes the ground back and the ground pushes the foot forward.
34. How much force would be required to accelerate a 5 kg ball at 3 m/s2?
5(3)=15 newtons.
35. A force of 150 N can accelerate a 10 kg mass at what rate?
a=(150)/10=15 m/s2.
36. Which has more momentum – a mass of A. 4 kg @ 3 m/s or B. 2.5 kg @ 5 m/s.
A. 4(3)=12N B. 2.5(5)= 12.5N. B has more momentum.
37. Explain how the same egg dropped from the same height breaks on a table but does not
break when it lands on a suspended blanket.
By increasing the time it in the collision, it decreases the force, explaining why an egg won’t break when contacting a blanket. On the table the impact time is very short causing a large force to exerted which breaks the egg.
38. Why is it easier to turn an empty shopping cart rather than a full one?
There is less mass which means less force is required to change its direction of motion.
39. Using , find the human reaction time if d = 28 cm.
0.24s
40. A car brakes from a speed of 20 km/h in a distance of 8 m. If the same car brakes from
a speed of 80 km/h what would the braking distance be?
80/20 = 4 so the speed factor is 4 which means the distance factor is 42 or 16. The braking distance would be 16(8m)= 112m
41. Complete the table.
V1 (km/h) / V2(km/h) / Initial Dist. Thrown (m)
From V1 / Velocity Factor =(V2/V1) / Distance Factor =(V2/V1)2 / Final Dist. Thrown =Init. Dist. x (V2/V1)2 From V2
10 / 20 / 5 / 2 / 4 / 20
20 / 40 / 15 / 2 / 4 / 60
15 / 60 / 18 / 4 / 16 / 288
30 / 90 / 5 / 3 / 9 / 45
25 / 125 / 18 / 5 / 25 / 450
12 / 72 / 6 / 6 / 36 / 216
40 / 40 / 6 / 1 / 1 / 6
42. Explain the three main mathematical relationships we modeled for grade 10 science
graphing them and labeling them.
InverseLinearQuadratic
43. An object is thrown 22 m from a vehicle going 15 m/s. How fast would the same vehicle
be going if the object was thrown 88 m? 88/22 = 4 so distance factor is 4 therefore the
speed factor is √4 or 2. So the speed is 15(2)= 30 m/s
- Define kinetic energy, potential energy, inertia, momentum and force.
Kinetic Energy- The kinetic energy of an object is the energy it possesses because of its motion. (Bowling Ball rolling)
Potential Energy- Potential energy is a type of energy an object has because of its position. (Boulder on a hill)
Inertia- Ability of an object to resist change in motion. (Train turning a sharp corner and tipping over)
Momentum- The quantity of motion of a moving body. (Boulder rolling down a hill)
Force- Strength or energy as an attribute of physical action or movement.
45. In terms of impulse and momentum, describe the motion of the following ticker tape
pattern.
…………… ......
acceleration occurs so an unbalanced force acts over some time (impulse)
46. A car travels an icy road (k=0.15) at 12 m/s. What is the braking distance?
d= 0.15(12)2 = 21.6m
47. Discuss how distractions can affect the reaction time of a driver.
The more distracted a driver is the more reaction time increases.
48. Complete the table for braking distance. d=kv2 (k= coefficient of friction)
Velocity (m/s) / Dry Pavement .06 / Wet Concrete .1 / Snow + Ice .155 / 1.5m / 2.5m / 3.73m
10 / 6m / 10m / 15m
15 / 13.5m / 22.5m / 33.75m
20 / 24m / 40m / 60m
25 / 37.5m / 62.5m / 93.75m
30 / 54m / 90m / 135m
40 / 96m / 160m / 240m
49. Complete the following table for total stopping distance with a reaction time of 1.5s.
d= kv2 + vt
Velocity (m/s) / Dry Pavement / Wet Concrete / Snow + Ice5 / 9m / 10m / 11.25m
10 / 21m / 25m / 30m
20 / 54m / 70m / 90m
30 / 99m / 135m / 180m
50. Describe how 3 safety features work on vehicles using Newton’s laws.
Seat belts- Is the unbalanced force that stops your inertia.
Crumple Zones- Reduces the force by increasing the time.
Air bags- Reduces the force by increasing the time.