Kinematics Examples
 A world class sprinter can
burst out of the blocks to
essentially top speed(of
about 11.5 m/s) in the first
15 m of the race.
 What is the average
acceleration of this
sprinter and how long
does it take to reach that
speed?
Kinematics Examples
 Read and Reread
 Draw Diagram w/axes
 List quantities
(known/unknown)
 Physics Principle
 Find an equation
 Estimate a reasonable answer
 Calculate and decide if
reasonable
Kinematics Examples
 Let's say you're driving in your car,
approaching a red light on Delaware
Avenue. A black Porsche is stopped at
the light in the right lane, but there's noone in the left lane, so you pull into the
left lane. You're traveling at 40 km/hr,
and when you're 15 meters from the stop
line the light turns green. You sail
through the green light at a constant
speed of 40 km/hr and pass the Porsche,
which accelerated from rest at a constant
rate of 3 m/s2 starting at the moment the
light turned green.
Kinematics Examples
 (a) How far from the stop line do
you pass the Porsche?
 (b) When does the Porsche pass
you?
 (c) If a police officer happens to
get you and the Porsche on the
radar gun at the instant the
Porsche passes you, will either of
you be pulled over for speeding?
Assume the speed limit is 50
km/hr.
Kinematics Examples
 Origin = stop line
Positive direction = the
direction you're traveling
Kinematics Examples
Kinematics Examples
 Step 2 - Figure out what you
need to solve for. At the
instant you pass the Porsche,
the x values (yours and the
Porsche's) have to be equal.
You're both the same
distance from the stop line,
in other words. Write out the
expression for your x-value
and the Porsche's.
Kinematics Examples
 We'll use the equation:
 x = xo + vot + 1/2 at2
 For you : x = -15 + 11.11 t +
0
 For the Porsche : x = 0 + 0 +
1/2 (3) t2 = 1.5 t2
Kinematics Examples
 At some time t, when you pass
the Porsche, these x values will
be the same. So, we can set the
equations equal to one another
and solve for time, and then
plug the time back in to either
x equation to get the distance
from the stop line. Doing this
gives:
 -15 + 11.11 t = 1.5 t2
Kinematics Examples
 Bringing everything to one
side gives:
 1.5 t2 - 11.11 t + 15 = 0
 This is a quadratic equation,
which we can solve using the
quadratic formula:
Kinematics Examples
 where a = 1.5, b = -11.11,
and c = 15
 This gives two values for t, t
= 1.776 s and t = 5.631 s.
Kinematics Examples
 What do these two values mean?
In many cases only one answer
will be relevant, and you'll have
to figure out which. In this case
both are relevant. The smaller
value is when you pass the
Porsche, while the larger one is
when the Porsche passes you
back.
Kinematics Examples
 To get the answer to question
(a), plug t = 1.776 into either
of your x expressions. They
should both give you the
same value for x, so you can
use one as a check
Kinematics Examples
 For you, at t = 1.776, x =
4.73 m.
For the Porsche, at t = 1.776
s, x = 4.73 m.
 We've actually already
calculated the answer to (b),
when the Porsche passes
you, which is at t = 5.6 s.
Kinematics Examples
 To get the answer to part (c),
we already know that you're
traveling at a constant speed
of 40 km/hr, which is under
the speed limit. To figure out
how fast the Porsche is going
at t = 5.631 seconds, use:
 v = vo + a t = 0 + (3) (5.631)
= 16.893 m/s.
Kinematics Examples
 Converting this to km/hr
gives a speed of 60.8 km/hr,
so the driver of the Porsche
is in danger of getting a
speeding ticket.