Transcript PreLecture 04

Lecture 04:
Kinematics + Dynamics

Kinematics Equations
constant acceleration

Dynamics
Newton’s Second Law
Non-zero acceleration
Equations for
Constant Acceleration
x
v
= x0 + v0t + ½ at2
= v0 + at
 v2
= v02 + 2a(x-x0)

x is final position

xo is initial position

v is final velocity

v0 is initial velocity

a is acceleration

t is time
Kinematics Example


A car is traveling 30 m/s and applies its breaks
to stop. Assuming constant acceleration of -6
m/s2, how long does it take for the car to stop,
and how far does it travel before stopping?

x = x0 + v0t + ½ at2

v = v0 + at

v2 = v02 + 2a(x-x0)
Begin by using the second
equation to find the time:

v = v0 + a t
Then use the first equation to find the
distance:
(0 m/s) = (30 m/s) + (-6 m/s2) t
t=5s
x = x0 + v0t + ½ a t2
x = (0 m) + (30 m/s)(5s) + ½ (-6 m/s2) (5 s)2
x = 75 m
Dynamics: F = ma

We have already dealt with
situations where a = 0.

But when the net force is not
zero, there IS an acceleration!
Dynamics Example

A tractor is pulling a trailer with a constant acceleration. If
the forward acceleration is 1.5 m/s2, Calculate the force on
the trailer (m = 400 kg) due to the tractor. (Consider just the
trailer.)
FBD:
x-direction:
FN
FT = ma
FT
Fg
F = ma
FT = 600 N
y-direction:
F = ma
FN - Fg = 0
y
x
FN = 3920 N
FT = 600 N
Summary of Concepts
• Constant Acceleration

x = x0 + v0t + ½ at2

v = v0 + at

v2 = v02 + 2a(x-x0)
• F = m a
– Draw Free Body Diagram
– Write down equations
– Solve
Kinematics Example

A car moving at 15 m/s is traveling toward an intersection
and sees the light turn yellow. The car accelerates at 4 m/s2
until it gets to the intersection 18 m away. How long does it
take the car to get to the intersection? (And assuming the
light is yellow for 1 s, does the car make it before the light
turns red?) Note: even after accelerating, the car is still traveling safely under the
speed limit…

There are two ways to solve this…I will use one
method and you will use the other!
Kinematics Example

A car moving at 15 m/s is traveling toward an intersection
and sees the light turn yellow. The car accelerates at 4 m/s2
until it gets to the intersection 18 m away. How long does it
take the car to get to the intersection? (And assuming the
light is yellow for 1 s, does the car make it before the light
turns red?) Note: even after accelerating, the car is still traveling safely under the
speed limit…

I will do this in two steps:
 First, using v2 = v02 + 2 a (x-x0), we can solve for the final velocity.
 Second, using v = v0 + a t, we can solve for the time.
Kinematics Example

A car moving at 15 m/s is traveling toward an intersection
and sees the light turn yellow. The car accelerates at 4 m/s2
until it gets to the intersection 18 m away. How long does it
take the car to get to the intersection? (And assuming the
light is yellow for 1 s, does the car make it before the light
turns red?) Note: even after accelerating, the car is still traveling safely under the
speed limit…
v2 = v02 + 2 a (x-x0)
v2 = (15 m/s)2 + 2 (4 m/s2) (18 m – 0 m)
v = 19.2 m/s
v = v0 + a t
(19.2 m/s) = (15 m/s) + (4 m/s2) t
t = 1.05 s
The car does not make it before the light turns red…
It should have just stopped!