Kinematics in a “Nutshell”
Vector Addition The notation for the last two questions is a bit different.. see pg. 185 (text).
1.
2.
3.
4.
5.
6.
7.
A soccer player leaves the bench and runs 25 m [N] and then 40 m [E]. Use a scale vector
diagram (on graph paper) to find the resultant displacement. (47m [N58°E])
Add the following vectors and determine the resultant: 2.0 m, [S50°E] and 4.0 m, [N25°E].
([4.0m N53°E])
Add the following vectors and determine the resultant: 4.0 m, [S35°W] and 4.0 m, [W15°S].
Use a scale vector diagram. (8.7m [S55°W])
Add the following vectors and determine the resultant. 3.0 m, [N45°W] and 5.0 m, [S35°E]
and 2.0 m, [N60°E]. (2.7m [E22°S])
Add the following vectors and determine the resultant: 6.0 m/s, [N25°E] and 2.0 m/s,
[E80°S]. (4.5m/s [N40°E])
Add the following vectors and determine the resultant: 3.0 m, 45° and 5.0 m, 135° and 2.0 m,
60°. (7.3m, 94°)
Subtract the following vectors and determine the resultant: 8.0 m/s, 330° and 4.0 m/s, 45°.
(7.9m/s, 300°)
Velocity
1.
2.
3.
4.
v
d
t
Remember, the slope of a position-time graph gives you the velocity.
A bicyclist travels at an average velocity of 11.2 km/h [W]. How far will the bicyclist travel in
175 minutes? (32.7 km)
A jogger runs 52 m [E] in 10.0s and then 41 m [E35°S] in 8.0 s. What is the jogger’s average
velocity? (9.8m/s [E17°S])
An air plane travels between two cities that are 2.4 x 103 km apart. The pilot wishes to follow
a ground course in the [N 45º W] direction. On the day of the flight this requires the pilot to
fly with a velocity of 2.0 x 102 km/h [N 35º W] and as a result the flight takes 8.0 h. What is
the wind velocity on that day? (111 km/h [N 62º W]).
A sports car is waiting for a red light to change. When the light turns green, a pickup truck
travelling at a constant speed passes the sports car as it starts to move.
The following graph represents the velocity of the two vehicles as a function of time.
v
(m/s)
20
Sports Car
15
Pickup Truck
10
5
0
1
2
3
4
5
6
7
At what time, t, will the sports car pass the pickup truck? (6s)
8
9
10 v (s)
5.
The graph below represents the velocity of a dandelion seed blowing in the wind for 20
seconds.
Velocity vs Time Graph
40
30
20
10
Velocity (m/s)
0
4
2
6
-10
8
10
12
14
16
18
20
Time (s)
-20
-30
-40
What is the displacement of the dandelion seed from the 4th second to the 20th second? (97m)
Acceleration a 
v
t
These problems are just about choosing the right equation to use from the
list below! The area under a v/t graph gives you the distance, the slope gives you the acceleration.
v f  vi  at
a)
b)
c)
2
2
1
x f  xi  vi t  a(t ) 2
2
1
x f  xi  (vi  v f )t
2
The following graph (below right) shows the velocity of a cannibal chasing his prey along a
foot bridge (e.g. straight line). Determine:
His displacement during the 8.0 s period.
(10m backward)
His instantaneous acceleration at point A.
(-5 m/s2)
Construct a position-time graph for the
cannibal in question 2.
Velocity-time graph of cannibal chasing prey
25
20
Velocity (m/s [foreward])
1.
v f  vi  2ax
15
10
5
0
-5 0
2
4
-10
-15
-20
-25
A
Time (s)
6
8
2.
3.
4.
5.
6.
7.
8.
9.
An Indy-500 race car's velocity increases from 4.00 m/s to 36.0 m/s over an interval
lasting 4.00 seconds. What is the car's average acceleration? How far does the car travel
during this amount of time? (8 m/s2 , 64m)
A golf ball rolls up a hill towards a mini-putt hole. It leaves the club traveling +2.0 m/s and
experiences an acceleration of - 0.50 m/s2. What will be its velocity after it has been
rolling for three seconds? (0.5 m/s2 )
An airplane starts from rest and accelerates at a constant 3.00 m/sec2 for 30.0 seconds
before leaving the ground at the end of the runway.
How long was the runway? (1350)
How fast was the plane moving when it "lifted off" the ground at the end of the runway?
(90m/s)
Over a period of 15 seconds a plane changes its velocity from 250km/h to 150 km/h at a
uniform rate of acceleration. How far does the plane travel during this amount of
time? (832m)
A snowmobile reaches a final velocity of 22.5 m/s after accelerating at 1.2 m/s2 from rest.
How long does this take? (18.75s)
A ball is dropped from the roof of a building. The ball takes 4.0 seconds to reach the
ground. Because of air friction, the ball falls with an average acceleration of 6.8 m/s 2.
What is the height of the building? (54.4m)
Free fall The key here is that the acceleration (due to gravity) is always -9.8m/s2. The velocity at
the highest point, vf, is 0. There are no horizontal values.
1. A student conducts a free-fall experiment while standing on the edge of the school roof
12.0 m above the ground. The student throws a heavy ball vertically upwards at a velocity of
5.00 m/s. On its way down, the ball narrowly misses the roof and hits the ground. How long
does it take the ball to hit the ground from the instant it is released?( 2.16s)
2. Laura propels a stone straight up with her slingshot. The stone strikes a tree branch 4.0 m
above her. It strikes the branch with a velocity of 5.0 m/s. How long does this take?
(0.53s)
3. A worker on a roof, 20.8 m above the ground, wants to drop a brick onto the back of a
moving truck. The truck is moving at a constant velocity of 5.0 m/s alongside the building.
The back of the truck is 1.2 m above the ground. How far from the point of impact must the
truck be, the instant the brick is dropped, so that the brick will land on the back of the
truck? (10m)
Projectile Motion For these problems you need to split up the velocity, displacement and
acceleration values into vertical and horizontal components, which are independent of each other. Use
the modified acceleration equations to do the problems. The horizontal velocity is constant (no
horizontal acceleration, vix = vfx ). The vertical acceleration is -9.8 m/s2. To find the angle of the
velocity vector hitting the ground, you find vfx and vfy, and use trig to find the resultant vector/angle.
x f  xi  vixt
1
y f  yi  (viy  v fy )t
2
v fy  viy  2ay
2
2
1
y f  yi  viy t  a y (t ) 2
2
y fy  yiy  at
Launched Horizontally.. There is no initial vertical velocity (viy).
1.
A softball rolls off of a 95cm tall table. It strikes the floor 80cm from the base of the
table. What is the velocity of the ball at the moment just before it hits the ground
(include direction)? (4.7m/s, 67o)
2.
A helicopter drops a relief package to some isolated refugees. The package travels 250
meters forward while taking 4.0 seconds to land after it is released.
a) How high was the helicopter flying when it released the package? (78.4m)
b) How fast was the package falling vertically just before it landed on the island? (-39.2m/s)
c) What was the package's impact speed, resultant velocity (with angle with respect to the
ground), when it landed on the island? (73m/s at 32o to the ground)
3. A pitcher throws a baseball horizontally with a velocity of 42m/s, releasing the ball from a
height of 2.0 m above the ground.
a) How long will it take for the ball to hit the ground? (0.63s.)
b) How far will the ball travel before hitting the ground? (27m)
c) What is the velocity of the ball at the moment just before it hits the ground (include
direction)? (42m/s at 8.4o to the ground)
Launched at an Angle.. Here, you can split the initial velocity into horizontal and vertical velocity.
Again, the vertical velocity, vfy , at the highest point is 0. The maximum range for a projectile is at a
launch angle of 45o.
1. A projectile has an initial horizontal component of velocity that is 72 km/h and an initial
vertical component of velocity that is 108 km/h. This means that the launch velocity is: (130
km/h at 56o)
2. A baseball is hit at an angle of 60o with a speed of 25.0 m/s. The horizontal component of
velocity when the ball is about to hit the ground is (12.5 m/s).
3. Yee-Ho is late for a balloon ride that is scheduled to leave from the bottom of a cliff. YeeHo just happens to be a good long jumper and takes off from the top of the cliff at the
precise moment when the balloon is taking off. The jumper leaves the cliff at 10 m/s at an
angle of 30o below the horizontal and the balloon rises straight up at 2.0 m/s. If the balloon
is 20 m from the base of the cliff when it starts and the jumper manages to land in the
basket of the balloon:
a. What is Yee-Ho's time in the air? (4 seconds)
b. What is the height of the cliff? (120m)
4. A water bomber, diving at an angle of 53° with the vertical, releases its water at an
altitude of 730 m. The water hits the fire (on the ground) 5.0 s after being released. How
far did the water travel horizontally during its flight? (treat the water as a projectile)
(1100m)
5. An object is launched from the ground at 40 m/s at an angle of 60 to the horizontal.
Determine how far the object has traveled when it has reached a height of 30 m above the
ground. (36m)
6. The Great Wall of China is approximately 25.0 m high. If a projectile is fired so that it will
achieve its maximum range, what is the slowest projectile that can be launched to just clear
the Great Wall? (ie at its maximum height) (31m/s)
7. A projectile launched at a 55 o angle will have the same range as a projectile launched (at a
35o angle)
8. A projectile is launched from the ground at 50 m/s at an angle of 65 o will have a maximum
height of: (100m)
9. A bullet is fired from a gun (at ground level) at an angle of 30 o with the ground. If the
muzzle speed of the gun is 48 m/s then velocity of the bullet upon striking the ground will
be: (-48 m/s, 330o )
10. What is the maximum range for a projectile launched at 30m/s? (92m)