Forces and Motion Kinematics

1. PHYSICS 1
KINEMATICS (OBJECTS IN MOTION)

2. SCALAR vs. VECTOR
 Scalar Quantities- Magnitude ONLY
 Example: Mass, Length, Distance, Speed
 5 meters
 Vector Quantities- Magnitude AND Direction
 Example: Weight, Displacement, Velocity,
Acceleration
 5 meters UP

3. DISTANCE and
DISPLACEMENT
 Distance (SCALAR)
 Displacement (VECTOR)
 Let’s say that you start from the bottom of the
arrow going UP which measures 4meters, and
the one going to the right is 3 meters
 DISTANCE (sum of the lines) = 3m+4m=7m
 DISPLACEMENT (length of diagonal line)=5m, by
Using Pythagorean Principle, is the two lines form
90 degrees, c^2=a^2+b^2

4. VELOCITY
 Velocity is the CHANGE in DISPLACEMENT
over TIME(CHANGE in TIME)
 2 Types of Velocity:
 Average Velocity
 Average Velocity=Total Velocity/Total Time
 Average Velocity=Vfinal+Vinitial/2
 Instantaneous Velocity
 Actual Value of Velocity at an exact point in time
 Can be obtained using the slope of displacement

5. ACCELERATION
 CHANGE in VELOCITY over TIME
 Acceleration=(Vfinal-Vinitial)/Time
 Acceleration= (+), causes increment/increase
in the value of velocity
 Deceleration=(-), cause decrement/decrease
in the value of velocity

6. DISPLACEMENT-TIME
GRAPH
 It can be seen that the object is placed a few
units away from the starting point and it moves
forward (can be seen as increase in the value of
displacement over time).
 A FLAT line in this graph means that the object is
AT REST or NOT MOVING.

7. VELOCITY-TIME GRAPH
 It can be observed that the value of velocity
INCREASE per unit time. This increase is due to
a POSITIVE value of ACCELERATION
experienced by the OBJECT. Similarly if the line
is going down it experiences NEGATIVE
ACCELERATION.
 Its initial value being (+) suggests that the object
is already moving at a (+) velocity at t=0.
 A flat line on this graph DOES NOT
NECESSARILY mean that the object is at
REST(except when flat at y=0), but means that
the body experiences CONSTANT
(UNIFORM/NON-CHANGING)VELOCITY.

8. KINEMATIC EQUATION
 If the object is moving with UNIFORM
MOTION (acceleration=0), (1) Distance =
Velocity*Time, ‘lest if acceleration is not 0, we
will use the other equations
Vfinal=Vinitial+a*t
Distance=Vinitial*t + 0.5a(t^2)
(Vfinal)^2-(Vinitial)^2=2a*distance
 Where, Vfinal is the final velocity, Vinitial is the
initial velociti and t is the time.

9. KINEMATIC EQUATIONS
 Use the APPROPRIATE EQUATION (Eqn 1
for a=0, and Eqns 2,3 and 4 if a is not equal to
0).
 Use equation that will lead to ONE
EQUATION-ONE UNKNOWN(what will be left
is just ONE UNKNOWN).
 Be careful of HIDDEN CLUES(indirectly stated
values), such as “going to stop”=> Vfinal=0, or
“at rest”=> Vinitial=0.

10. PRACTICE PROBLEMS
 Example 1: A car moving at 20m/s accelerates
at the rate of 2m/s^2 for 5 seconds. Find the
new velocity of the car.
 Answer: Vfinal = Vinitial + at
 Vfinal = 20m/s+(2m/s^2)*5s
= 30m/s

11.  Example 2: How long will it take a car moving
at 50m/s to stop if on the brake manufacturer’s
manual says that the breaks can cause
deceleration at the rate of 5m/s^2?
(deceleration is - acceleration)
 Answer: Vf=Vi+at, Vf-Vi=at, TIME=(Vf-Vi)/a
 Vf=0(STOP), TIME=(0-50m/s)/(-5m/s^2)
 TIME=10seconds

12.  Example 3: In Example 2, how long will be the
skid marks due to the stop? (Skid marks-
>distance before stopping)
Distance=Vinitial*t + 0.5a(t^2)
 Distance= (50m/s)*10s + 0.5(-5m/s^2)(10s)^2
= 500m + (-250m)
= 250m

13. PROJECTILE MOTION
 Similar to Kinematic Equations but involves
motion along BOTH the x and y axes.
 Motion along y is guided by
GRAVITY(acceleration=g=-9.81m/s^2, or -
32ft/s^2).
 Motion along x is UNIFORM, (acceleration=0).

14. PROJECTILE MOTION (Type
1)
 Occurs when an object is thrown
HORIZONTALLY from a given height.
 The height from which it was thrown is the
MAXIMUM HEIGHT REACHED by the
projectile.
 Even if along x, we can use Dx=Vx*t, (Dx is
the displacement along x or RANGE, and Vx is
the horizontal component of the velocity), we
neither have Dx nor the value of t, so we start
with values along y.

15. PROJECTILE MOTION (Type
1)
 Along y, we can use Dy(max height) = Vinitial
y*t + 0.5at^2
 Unfortunately if an object is thrown
HORIZONTALLY, there is no y component for
the initial velocity thus Vinitial y =0
 This leaves us with Dy = 0.5at^2
 2Dy/a = t^2
 Sqrt(2Dy/a) = t

16. PROJECTILE MOTION (Type
1)
 Example: Jr throws a ball horizontally, with
velocity of 5m/s from a building with
height=100m. If Melvin, his playmate has to
catch the ball as it falls down, neglecting air
resistance(and possible pranks used by Jr),
Find: (a)time in seconds, in which the ball will
fall into the ground and (b)the distance along
x, from the building, in which the ball will land.

17. PROJECTILE MOTION (Type
1)
 Answer: TIME = Sqrt (2Dy/g) =
Sqrt(2(100)/9.81)
 TIME = 4.52 seconds
 Distance along x(Range) = Vx*t
 Range = 5m/s*(4.52s) = 22.58m

18. PROJECTILE MOTION (Type
2)
 In this type of projectile, both the x and y will
have components of the initial velocity.
 If the angle formed is along the HORIZONTAL, the
value of Vx initial = Vinitial * cos(angle)
Vy initial = Vinitial * sin(angle)
It can also be noted that the MAXIMUM HEIGHT is
attained when Vy is set to 0. Initially the projectile
GOES UP (+velocity), then it GOES DOWN (-velocity),
and thus at MAXIMUM HEIGHT(middle), the value of
VELOCITY along y is EQUAL TO ZERO.

19. Projectile Motion (Type 2)
 Example: Jr and Melvin are still playing catch
ball and for this time, it is Jr who is to catch the
ball. If Melvin gives the ball an initial velocity of
10m/s at an angle 30degrees from the
horizontal, (a) what is the maximum height
attained by the ball? (b)how long is the ball’s
total time of flight? (c)what is the value of its
range(distance along x)?
 Answer:
 Vx initial=10*cos30 = 8.66m/s
 Vy initial=10sin30 = 5m/s

20. Projectile Motion (Type 2)
 Vfinal y = Vinitial y + at
 We set the Vfinal y to 0 to get t when the ball
attains its max height
 0=5m/s + -9,81*t
 -5= -9.81t
 T(max height)=0.51seconds
 Tflight=2T(max height)=1.02seconds

21.  Max height=Dy
 Dy=Vinitial y*T(max height)+0.5a[T(max
height)]^2
 Dy=5m/s*(0.51s)+0.5(9.81m/s^2)*(0.51) = 5.05m
 Dx=Vx initial*Tflight=8.66m/s*(1.02s)=8.83m
 NOTE: We use T(max height) to get Dy because,
Dy or the max height is attained at T(max height).
And we use Tflight for Dx, because the range
occurs when the ball falls to the ground, which is
attained at Tflight.