describing motion kinematics in one dimension and graphing motion kinematics worksheet answers
Dr.SamuelHunt,United Arab Emirates,Teacher
Describing Motion: Kinematics
in One Dimension(previous lecture)
• Reference Frames and Displacement
• Average Velocity
• Instantaneous Velocity
• Motion at Constant Acceleration
• Falling ObjectsRecalling Recalling Last Lecture Last Lecture
You need to defi ne a reference frame in order to fully characterize the motion of an
object. In general, we use Earth as reference frame for
Displacement is how far an object is
found from its initial position after an interval
of time Δt:
Eq. 2.1 gives the magnitude of the vector
Average velocity is given by:
And average acceleration is:
Both velocity and acceleration are vectors (give direction and magnitude) : Acceleration, velocity, displacement, and time.DecelerationThe car from the previous example, now moving to the left and
decelerating. Few Few Relevant Relevant Notes Notes
I. The magnitudes of the instantaneous velocity and instantaneous speed are the
same, though this is not necessarily true for average velocity and average s peed.
II. If I give you an interval of time Δt = t – t but do not specify the initial and final
time t and t , you can then assume t = 0 such that Δt = t – t = t = t.
1 2 1 2 1 2
III. Similarly, if I give you a displacement Δx = x – x but x is not specified, you can
2 1 1
assume x = 0 such that Δx = x – x = x = x.
1 2 1 2
II and III are just a matter of redefining the origin of your coordinate system. For
Translate the origin from “0” to “x ” by applying a transformation:
X = x – x such that
X = 0 – x and X = x – x are the new set of coordinates.
1 1 2 2 1Motion at Constant Acceleration Motion at Constant Acceleration
It is not unusual to have situations where the acceleration is constant. In this lecture,
we will see one of such situations experienced by you in your everyday l ife.
But before, let’s assume a motion in a straight line subject to a constant acceleration.
In this case the following is valid:
à The instantaneous acceleration and the average acceleration have the same
magnitude. We can then write:
We will use item II from the previous slide with following conventions:
t = 0 , t = t
and also define the position and velocity at t = t = 0 as
x = x , x = x
1 0 2
v = v , v = v
1 0 2
for our future developments. Motion at Constant Acceleration Motion at Constant Acceleration
With this in mind, let’s find some useful equations relating a, t, v, v , x and x for
cases with constant acceleration such that you can calculate any unknown variable if
you know the others .
Let’s first obtain an equation to calculate the velocity v of an object subjected to a
constant acceleration a:
Using the definitions introduced on the previous slide, we can rewrite eq. 2.6 as:
Eq. 2.10 can be rearranged:
Eq. 2.10 allows you to calculate the velocity v of an object at a given time t knowing
its acceleration a and initial velocity v .
0Motion at Constant Acceleration Motion at Constant Acceleration
Example: A car accelerates from rest at a constant rate of 5.0 m/s . What is its
velocity after 5.0 s?
v = 0.0 m/s
a = 5.0 m/s
t = 5.0 s
Using eq. 2.10:Motion at Constant Acceleration Motion at Constant Acceleration
We can also obtain an equation that gives the position x of an object knowing its
constant acceleration a, initial position x and initial velocity v :
Using our definitions, eq. 2.4 can be written as:
On the other hand, since the velocity of the object changes from v to v at a constant
rate a, we have:
(2.12)Motion at Constant Acceleration Motion at Constant Acceleration
But (Eq. 2.11) = (Eq. 2.12) . Then:
Using equation 2.10 in the above expres sion:
Eq. 2.13 allows you to calculate the position of an object at a given time t knowing
its acceleration a , initial velocity v and initial position x .
0 0Motion at Constant Acceleration Motion at Constant Acceleration
There is another very useful equation that applies to motion at constant acceleration
in situations where no time information has been provided.
We will use equations 2.10 and 2.13 to el iminate the time variable:
From eq. 2.10 à
Using this expression in 2.13:
(2.14)Motion at Constant Acceleration Motion at Constant Acceleration
Example (Problem 28 from the text book): Determine the stopping distances for a
car with an initial speed of 95 Km/h and human reaction time of 1.0 s, for an
acceleration (a) a = -4.0 m/s ; (b) a= -8.8 m/s ?
We want to convert from h to s. It is also wise to convert from Km to m:
(remember to c arry some extra figures
through your calculations)
Driver realizes he has to brake (v = v )
Driver starts braking (v = v )
Next, we have to define our
reference frame (coordinate system)
in order to calculate the position
where the driver starts breaking.
cars stops (v = 0)Motion at Constant Acceleration Motion at Constant Acceleration
Note that no information on the time needed to stop the car is provided. This problem
is better approac hed with eq. 2.14. The position where the brak es are applied can be
obtained if we observe that the car is moving at constant velocity v before the driver
reacts. We can then use the equation of motion for constant velocity to find x :
As we have already mentioned, the final velocity v is zero in both (a) and (b) c ases:
v (final) = 0.0 m/s
(continue on the next slide)Motion at Constant Acceleration Motion at Constant Acceleration
We have then the fol lowing initial conditions:
(a) The stopping distance for a = -4.0 m/s is:
(b) Similarly, the stopping distance for a = -8.0 m/s is:Motion at Constant Acceleration Motion at Constant Acceleration
Let’s summarize the set equations we just found:
Note that these equations are only valid when the acceleration is constant.Motion at Constant Acceleration Motion at Constant Acceleration
Example (Problem 21 from the text book): A car accelerates from 13 m/s to
25 m/s in 6.0 s. (a) What was its acceleration? (b) How far did it travel in this time?
Assume constant acceleration.
This problem is about choosing an appropriate equation from the set of equations we
just found. For this purpose, first you have to write the parameters given in the
v = 13 m/s
v = 25 m/s
t = 6.0 s
(continue on the next slides)Motion at Constant Acceleration Motion at Constant Acceleration
(a) Here you want a knowing v , v and t. Which of the equations below you find more
appropriate to solve this problem?
It is clear that 2.10 is the better choice for this item.
;Motion at Constant Acceleration Motion at Constant Acceleration
(b) Here you want the di stance travelled (x – x ) knowing v , v , t and a (calculated
in item (a) ). Now, which of the equati ons below you find more appropri ate to solve
Either equation 2.12 or 2.13 is a good choice. Let’s use 2.12: