# Kinematics

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```					General Uses
   These are used to describe the
relationships between the following
kinematic quantities:
○ Distance/displacement
○ Speed/velocity
○ Time
○ Acceleration
   When there is an unknown, it can be
solved for when the values of the other
quantities are given
The Four Basic Kinematic
Equations are:

   V = V0 + a Δt

   V2 = V02 + 2aΔs

   S = V0Δt + 0.5 a Δt2

   S = (V0 + V)/2 × t
V = V0 + a Δt

E.g. A car starts at rest and accelerates
uniformly at 2 m/s2 for 5 seconds and
stops accelerating from here on.
Calculate its velocity after t = 5 seconds.

Using V = V0 + a Δt, we sub in values 0 for
V0, 2 for a and 5 for t. Solving for V, we
get:
V = 10 m/s
V2 = V02 + 2aΔs

E.g. A train accelerates from 10 m/s to 40
m/s at an acceleration of 1m/s 2. what
distance does it cover during this time.

Using V2 = V02 + 2aΔs, we sub in values
40 for V, 10 for V0 and 1 for a. Re-
arranging to solve for s, we get:

S = 750 m
S = V0Δt + 0.5 a Δt2

E.g. A body starts from rest at a uniform
acceleration of 3 m/s2. how long does it
take to cover a distance of 100m.

Using S = V0Δt + 0.5 a Δt2, we sub in values 3
for a, 0 for V0 and 100 for s. Re-arranging
the equation and solving for t (using the
t = 8.51 or -8.51 seconds. As time cannot be
negative, t = 8.51 seconds.
S = (V0 + V)/2 × t

A car decelerates from 20 m/s to 10 m/s
over a period of 10 seconds. How far
does it travel during this time period.

Using S = (V0 + V)/2 × t, we sub in values
20 for V0, 10 for V and 10 for t. Solving
for s, we get:
S = 150m
Note:
   All units must be converted such that
they are uniform for different variable
throughout the calculations.

   Kinematic quantities that are scalar
CANNOT be negative, hence any such
alternate solutions obtained must be
disregarded.
Standard units for the various
quantities are as follows:
   Speed – metres/second

   Acceleration – metres/second squared

   Distance – metres

   Time - seconds

```
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 views: 13 posted: 10/23/2011 language: English pages: 10