# Kinematics Problems and Equations

```Mechanics: The study of motion of objects ;]
 Kinematics… WHAT IS THAT?

› The science of describing the motion of objects
 Measure by using graphs, diagrams, numbers, and
equations
Scalars- quantities that are fully explained by
magnitude alone
 Vectors- quantities that are explained by
magnitude and direction

› THE SIGN MATTERS!
The sign matters b/c it determines the
direction of the object
 Distance and displacement both show how an
object moves

› Distance is a scalar and shows the amount of
distance covered
› Displacement is a vector and shows the overall
change in an objects position

Speed and velocity are both important to
know in order to understand kinematics

In some instances, you may not know a variable of
an object’s motion and must solve for it!
› A car has a velocity of 23 m/s East; and has an
acceleration of 9 m/s2. What don’t we know?
› Use the equations:
1 2
x  vi t  at
2
(v f  vi )
2
x 
2a
2
v f  vi  at
 vi  v 2 
x  
t
 2 
Make a list of givens
 Identify what needs to be found
 Jot what equation will be used
 This will limit your chances of making error on
 v= velocity; a=acceleration; t= time ,
Δx
=displacement



Ima Hurryin is approaching a stoplight moving with a
velocity of +30.0 m/s. The light turns yellow, and Ima
applies the brakes and skids to a stop. If Ima's
acceleration is -8.00 m/s2, then determine the
displacement of the car during the skidding process.
To prepare for solving we would make a list:
› Vi = + 30.0 m=/s
› Vf = O m/s
› a= -8 m/s2
› x = ?
Make note of
negative and
positive signs!

Make a picture:
What equation
should we use?


Solve…
(v f  vi )
2
› Equation: x 
2
2a
(0  30 )
x 
= 56.25 m
2(8)
2
2
 Ben Rushin is waiting at a stoplight. When it finally
turns green, Ben accelerated from rest at a rate of a
6.00 m/s2 for a time of 4.10 seconds. Determine the
displacement of Ben's car during this time period.

List:
› Vi = 0 m/s
› t= 4.10 s
› a= 6 m/s2
› d= ?

Picture:

Equation: x  v t  1 at 2
i
2

1
2
x  (0)( 4.10)  (6)( 4.10) = 50.43 m
2
Free Fall






Falling under the sole influence of gravity
Downward acceleration of 9.8 m/s/s
The motion of an object in free fall can be described
by kinematics equations
If an object is dropped from an elevated height its
initial velocity is zero
If an object is projected upwards in a vertical
direction it will slow down as it rises. When its peak
is reached the velocity is 0 m/s.
If an object is projected upwards in a vertical
direction the velocity at which its projected is equal in
magnitude and opposite in direction to the velocity it
has when it returns to the original height
Example 3

Luke Autbeloe drops a pile of roof shingles
from the top of a roof located 8.52 meters
above the ground. Determine the time
required for the shingles to reach the ground.

To prepare for solving we would make a list:
› Vi = 0 m=/s
› d = -8.52 m
› a= -9.8 m/s2
› t= ?
Example 3 Continued…

Make a picture:
What equation
should we use?

Example 3 Continued…
 Solve…
› Equation:
 -8.52 m = (0 m/s)*(t) + 0.5*(-9.8
m/s2)*(t) 2
› Solution:
 t= 1.32 s
Example 4 Continued…

A race car accelerates uniformly from 18.5
m/s to 46.1 m/s in 2.47 seconds. Determine
the acceleration of the car and the distance
traveled.

To prepare for solving we would make a list:
› Vi = 18.5 m=/s
› Vf = 46.1 m/s
› t= 2.47 s
› d= ?
› a= ?
Example 4 Continued…
a = (delta v)/t
a = (46.1 m/s - 18.5 m/s)/(2.47 s)
a = 11.2 m/s2
d = vi*t + 0.5*a*t2
d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s2)*(2.47
s) 2
d = 45.7 m + 34.1 m
d = 79.8 m
```