Brooklyn College 1

One & two dimensional motions with constant acceleration

Purpose

1. To be able to identify a free fall motion.

2. To be able to use the kinematic equations.

3. To study two dimensional motion with constant acceleration.

Introduction

If a quantity has a constant value, then the average of this quantity is equal to this constant value. For example the

average of 4, 4, and 4 is (4+4+4)/3= 4. The velocity is defined as the time rate of change of position. Its unit is m/s. The

average velocity is defined as

, where , is the final position, is the initial position and is the

displacement of the object. The velocity has direction and numerical value. We call the numerical value the magnitude.

The magnitude of the velocity at each instant is the speed at that instant. If an object has a constant velocity, , then

, so

, we will write ‘t’ to represent ‘t’, the time of motion.

Solving for , we get eqn. (1). Equation 1 is valid only for motion with constant velocity, because we

used the condition that is constant so that to derive it.

If the velocity of an object is changing, then the object has acceleration. The acceleration is defined as the time rate of

change of velocity. Its unit is m/s2. The average acceleration is defined as

, where is the final velocity,

is the initial velocity, and t = t, is the time of the motion.

We will now derive two important equations for the case of constant acceleration.

Similar to the argument above, if the acceleration is constant, then this constant acceleration,

. Then

solving for , we get:

eqn. (2).

Equation 2 is a straight line equation for as a function of . Therefore, for this case,

, the middle point. See

figure 1.

But from the definition of the average velocity

.

Using these 2 expressions for , we get:

. Then solving for

we get:

.

Substitute this into eqn. 2 and solving for x, we get:

,

but , therefore

eqn. (3).

to tfinal t t

v

Figure 1: For constant acceleration,

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Equations 2 & 3 are valid only for motion with constant acceleration. They are called the kinematics equations. They can

be applied to any direction (horizontal or vertical, or other) where the acceleration, is constant.

Notice that if is constant and , eqn. 3 is a parabola for the position, as a function of .

If an object is under no other influences expect the attraction of the earth, “gravity”, the object motion is said to be in a

“free fall” motion. In this case the acceleration is constant and equal to – (if we consider the downwards direction to

be –ve), where (recall experiment 2: Measurements and acceleration due to gravity).

Let’s capture some of these concepts using a simulator.

Open the simulator: http://physics.bu.edu/~duffy/HTML5/1Dmotion_constantv_constanta.html

The red car moves with constant velocity, but the blue car moves with constant acceleration. Set the initial position of the

red car to 0 m, and set the initial (and constant velocity) of the red car to 5 m/s. Set the constant acceleration of the blue

car to 1.3 m/s2.

i) For the red car, what do you expect the velocity graph to be? Choose an answer: (a) Horizontal. (b) Straight line with

+ve slope.

ii) For the blue car, what do you expect the velocity graph to be? Choose an answer from the same answer choices.

Now click at the bottom of the simulator Graph “Velocity vs. time”. Click play, and keep it playing till it stops. Verify your

answer for the velocity graphs.

iii) What does the intersection point of the graphs represent?

iv) For the red car, what do you expect the position versus time graph to be? Choose an answer: (a) Horizontal, (b) A part

of a parabola, (c) Straight line with +ve slope.

v) For the blue car, what do you expect the position versus time graph to be? Choose an answer from the same answer

choices.

Now click at the bottom of the simulator Graph “position vs. time”. Click Reset then click play, and keep it playing till it

ends. Verify your answer for the position graphs.

The data sheets are on pages 6 and 7

Part 1: Free fall from rest

1) Open the simulator https://www.compadre.org/Physlets/mechanics/illustration2_6.cfm

read the explanatory text for animation 1. Then click animation 1 and play till it ends. Observe the graphs.

2) Right click the acceleration graph and expand it by dragging from the lower right corner. If you click with the mouth at a

graph point, the simulator displays its (x, y) coordinates. Here the x-axis represents time. Click with the mouse on a point

of the graph where time = 2 s, to see the value of acceleration, .

a) Record the values of (including its sign), at t = 2 s and at t = 3 s.

b) Is the acceleration, constant? c) What is its value? d) Accordingly is the motion a free fall motion?

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3) Close the acceleration graph. Right click the velocity graph, and expand it.

a) What is the initial value of the velocity? b) Why does the initial value of the velocity has this value?

Click on any 2 points of the graph and calculate the slope: rise / run.

c) What is the value of the slope?

d) How does it relate to the value of the acceleration a? (Hint: See eqn. 2, given in the introduction). Close the velocity

graph.

4) Notice the graph for position.

a) Is the parabolic path for the position graph open up or down? b) Why? (Hint: See eqn. 3, given in the introduction).

Part 2: Free fall with initial

1) In the same simulator of part 1, click animation 2. Read the explanatory text for animation 2. Here we are going to

examine the motion of an object thrown upwards from the ground with an upwards initial velocity, . Click play. Keep it

playing till it ends.

2)

a) Is the whole motion for animation 2 considered a free fall motion?

b) Why? (Hint: See the term free fall mentioned in the introduction).

c) Is the acceleration, constant? d) What is its value?

3) Notice the velocity graph.

a) What is the value of ?

b) Why isn’t it zero? (Hint: See the explanatory text for animation 2).

The maximum height is the peak of the y-position graph.

c) Using the simulator graph, what is the value of at the time when the position is at maximum height?

d) Using the value of and t, write an expression for as a function of t? (Hint: See eqn. 2 given in the introduction).

e) Now for t = 2.5 s, calculate , using your expression, (show your work). Record the value.

Now verify using the graph of by clicking on the point on the velocity graph where t= 2.5 s. Was your answer to e)

correct? If not, try again.

4) Now notice the position graph.

a) What are the values of the initial position and the final position? Examine back again the velocity graph.

b) What is the value of at the final time?

c) How does it relate to the value of ?

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d) Is the position graph symmetrical for the motion where the magnitudes of and ?

e) In this case, how does t of maximum height relate to t final?

f) Using the values of and , write an expression for as a function of . (Hint: Is the acceleration constant? See

eqn. 3 given in the introduction).

g) Now using your expression calculate at t = 2.5 s. Verify your answer using the position graph of the simulator, by

clicking the position graph at the point where t= 2.5 s and observing the value of the position displayed.

h) Did you get the correct answer? If not, try again.

5) For = 3 m, = 1.9 m/s, and the constant acceleration = - 2 m/s2 (notice the negative sign of the acceleration ):

a) Write an eqn. for as a function of , and an eqn. for as a function of . (Hint: See eqns. 2 & 3 given in the

introduction).

Now open the simulator: http://ophysics.com/k6.html Set the values just mentioned here at the beginning of step 5.

Notice & record the displayed equations by the simulator for and , verify your answer.

b) Using the equation for , calculate for the maximum +ve value of position. With the simulator paused, verify this

value of from the simulator by dragging the time dot to that value of you just calculated and noticing the graphs of

and .

c) Was your answer correct? If not, try again.

Part 3: Two-dimensional projectile motion

1) In the introduction, it was mentioned that equations 2 & 3 are valid for any direction where the acceleration is

constant. Also it was mentioned that the speed is the magnitude of the

velocity at each instant.

We will examine the case when an object in an x-y plane (horizontal-

vertical plane) is projected with a launch angle , and with a launch

speed , see figure 2.

2) The initial velocity (has direction) as shown, makes an angle o with

the x-axis. The x-component of the initial velocity has value

eqn. (4).

Similarly, for the y-component

eqn. (5).

The speed

, Why? (Hint: Pythagoras theorem).

The range, is defined for a given projectile motion, as the maximum horizontal distance in the x-direction reached by

the projectile. See figure 2.

o

x

y

Figure 2: A projectile with initial speed vo

and initial angle o.

h max

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3) Now let’s examine the acceleration. As the object is fired, and in the air, and if we neglect air resistance, the only force

is gravity, so, as mentioned in the introduction, the object is in a free fall motion. So if we take the downward direction as

negative, the acceleration, . But notice that gravity is attraction vertically down towards the

center of the earth, so the acceleration of free fall is the ‘y-component’ of the acceleration, .

In the x-direction, there are no influences (we are neglecting air resistance), so there is no acceleration in the x-direction,

(still a constant but equal to zero). Therefore we can use equations 2 & 3 separately in each of the y-

direction and the x-direction (also equation 1: constant velocity applies to the x-direction, do you know why? (Hint: see

the definition of the acceleration given in the introduction)).

4) If and , if we calculate and using eqns. 4 & 5 given in step 2 above, we

find that . Now let’s consider the motion of the projectile from up to the time when

final becomes again. See fig. 2.

We analyze the motion in the y-direction and in the x-direction. It is very important to notice that the time variable is

common to both directions.

In the y-direction: .

a) For final to become equal to : Using eqn. 3 given in the introduction, calculate final.

b) Using eqn. 2 given in the introduction, calculate final.

c) Using eqn. 2, calculate the time for reaching maximum height. (Hint: what is the value of at the maximum height?).

In the x-direction: (does change? Hint: see the question and hint in step 3 above),

.

d) Using your answer for final in point (a) above for the y-direction, and using equation 1, given in the introduction,

calculate the range, for the motion up to final .

5) Open the simulator http://ophysics.com/k8.html keep all default settings. Notice the default settings match the motion

we considered in step 4 above. Click fire. Pause when y final becomes about 0 m (after you pause you can fine tune using

click and drag for the time setting dot and then using the up step and down step ).

6) a) Is the position graph symmetrical? Note that we are neglecting air resistance and considering the motion up to

final = .

For the time, you have set in the pause mode, the simulator values, at that time, , for , and are displayed

under the graph.

Now verify your answers to the calculations of step 4 [(a) final, b) final, c) for maximum height, and d) range ].

Did you get the right answers? If not, try again.

7) Repeat steps 4, 5 and 6 for and

.

8) Which of the 3 trials ( or ) gives the maximum range?

9) Open the simulator http://physics.bu.edu/~duffy/HTML5/projectile_motion_spray.html

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Keep all the default settings. Click play.

a) Which gives the largest maximum height?

b) Which gives the maximum range?

Data sheet

Introduction exercise:

Answer for i) Red car velocity graph: Answer for ii) Blue car velocity graph:

Answer for iii) Intersection point:

Answer for iv) Red car position graph: Answer for v) Blue car position graph:

Part 1: Free fall from rest

Step 2)

at t = 2 s at t = 3 s Is constant? value of Is the motion free fall?

Step 3)

Reason for value of value of slope slope relation to acceleration

Step 4) Answer for a) Answer for b)

Part 2: Free fall with initial

Step 2)

Is the whole motion considered free fall?

Reason Is constant? value of

Step 3)

value of Reason for value of at maximum height as a function of t at t = 2.5 s

Step 4)

at final time at final time compare at final time to Is the position graph symmetrical?

In this case, how does the time of maximum height relate to the final time?

f) Expression for as a function of : g) Calculate at = 2.5 s: h) answer:

Step 5) a) as a function of : as a function of :

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Equations for & displayed by simulator:

Calculated value of at maximum height Measured value of at maximum height Was your calculated value correct?

Part 3: Two-dimensional projectile motion

Step 4)

a) Calculate final:

b) Calculate final:

c) Calculate the time for reaching maximum height:

d) Calculate the range for the motion up to final :

Step 6) a) Is the position graph symmetrical?

measured final measured final measured for maximum height

measured range

Step 7)

a) Calculate final:

b) Calculate final:

c) Calculate the time for reaching maximum height:

d) Calculate the range for the motion up to final :

For : Repeated step 6) a) Is the position graph symmetrical?

measured final measured final measured for maximum height

measured range

a) Calculate final:

b) Calculate final:

c) Calculate the time for reaching maximum height:

d) Calculate the range for the motion up to final :

For : Repeated step 6) a) Is the position graph symmetrical?

measured final measured final measured for maximum height

measured range

Step 8) Max Range

is for