Experiment 4
Uniform velocity and uniformly accelerated motion
In this laboratory exercise you will learn the kinematics of an object that is
subjected to (i) uniform velocity and (ii) uniform acceleration i.e. an object moving with
a uniform or constant velocity or acceleration. The most common example of an object
moving with a constant acceleration is a free falling object. If the object falls solely under
the influence of gravity it is said to be in free fall. In this case the acceleration of the
object is the acceleration due to gravity g. Near the Earth’s surface the acceleration of
gravity is essentially constant with a value of
g = 9.80 m/s2 = 980 cm/s2 = 32.2 ft/s2
(1)
Air resistance can affect the acceleration of a freely falling object. However, for
relatively heavy objects this effect can be negligible. In this experiment we will use two
methods to measure the acceleration due to gravity. In one case we will use a motion
sensor and in the other case we will use a photogate timer.
Objectives:
1. Measure the position and velocity of an object that is moving with (i) a constant
velocity and (ii) constant acceleration.
2. Explore the relationships between position, velocity and acceleration.
3. Analyze the graphs of position and velocity vs. time for a cart’s motion.
4. Explain how the slope of the velocity graph relates to the acceleration of the
object.
5. Analyze the motion of an object in free fall.
Theory:
Horizontal and Inclined plane
The purpose of this experiment is to study the relationships between position, velocity
and acceleration in linear motion when the object moves with (i) a constant velocity
(horizontal plane) and (ii) with constant acceleration (inclined plane).
Motion sensor
Cart
Figure 1. Cart moving away from the sensor with a constant velocity
As seen in Figure 1, the cart rests on a frictionless horizontal plane. When the cart is
given a slight push, it moves with a constant velocity. Using the motion sensor one can
1
graphically plot the position versus time, velocity versus time and acceleration versus
time of the moving cart.
The velocity at any instant (v) of an object is obtained from the average velocity by
shrinking the time interval ∆t to zero. As ∆t approaches zero, the average velocity
approaches the instantaneous velocity. This is written in mathematical form as follows:
v = ∆t lim
→ 0
∆x dx
=
∆t dt
(2)
The equation above states that the instantaneous velocity of the object is the derivative of
the position of the object with respect to time. Thus the slope of the objects position-time
curve at any instant of time gives the instantaneous velocity of the object at that time.
When an objects velocity changes with time, the object is said to undergo acceleration a.
For motion along an axis, the average acceleration aavg over a time interval ∆t is given by:
a avg =
v 2 − v1 ∆v
=
t 2 − t1
∆t
dv
dt
dv d dx
d 2x
a=
= ( )= 2
dt dt dt
dt
a=
(3)
where the object has velocity v1 at time t1 and velocity v2 at time t2. The instantaneous
acceleration is the derivative of the velocity with respect to time. From Equations 2 and 3
it is seen that if the object moves in a straight line with a constant velocity the
acceleration is zero.
Motion sensor
N
Cart
mgSin(θ)
mgCos(θ)
mg
θ
Figure 2. Cart moving down an inclined plane with constant acceleration. The
of the force along the plane is mgSin(θ).
2
Figure 2 shows a cart moving with a constant acceleration down a frictionless inclined
plane. The only force acting on the cart is the force due to gravity. The components of
this force are also indicated in Figure 2. Applying Newton’s second law of motion to the
cart, we can get two equations along the direction of motion and perpendicular to it.
These equations are given below:
∑ F = ma
Along the plane we have
mgSin(θ ) = ma
Perpendicular to the plane we have
mgCos(θ ) − N = 0
(4)
Using the motion sensor one can find the acceleration of the cart down the inclined plane
and hence calculate the acceleration due to gravity by finding the angle of inclination of
the plane and by using Equations 4. Compare this value of g with that given in Equation 1
above.
Free fall
An object in free fall (neglecting air resistance) falls under the influence of gravity, or
with an acceleration g. The distance y the object falls in a time t is given by:
1
y = y 0 + vo t + gt 2
(5)
2
where yo is the initial position and vo is the
initial velocity. If an object is dropped
from rest (vo=0) and from the origin
(yo=0), then
y=
1 2
gt
2
y
(6)
(downward taken as positive). Hence by
measuring the time t is takes for an object
to fall a distance y, the acceleration due to
gravity can be easily calculated.
Figure 3. Free fall apparatus for determining
the acceleration due to gravity.
In this part of the experiment you will use a photogate timer to measure the time of fall of
a steel ball for different heights “y” as seen in Figure 3. Fill in the data of the height and
the time of fall in Table 1. Take at least ten different heights. Since the ball is always
dropped from rest we can use Equation 6 to find g. Complete Table 1 by calculating t2.
Then make a plot of distance fallen “y” versus “t2”. The slope of this straight line will be
the acceleration due to gravity. Compare this value to that given in Equation 1.
3
Exercises
1. If the angle of inclination θ is 15o in Figure 2, what is the acceleration of the cart
if the cart is released from rest at the top of the plane? Assume g = 9.8 m/s2.
Solution
Using Equations 4 we have: mgSin(θ) = ma
a = gSin(θ) = 9.8 Sin(15o) = 2.54 m/s2
or
2. A frightened ostrich moves in a straight line with a velocity described by the
velocity-time graph shown below. Sketch the acceleration versus time graph.
Velocity (m/s)
4
2
0
-2
-4
0
50
100
150
200
Time (s)
Solution
acceleration (m/s2)
4
2
0
-2
-4
0
50
100
150
200
Time (s)
4
3. How far does the runner whose velocity-time graph is shown in the Figure below
travel in 16 s.
10
B
Velocity (m/s)
8
C
6
4
D
A
2
0
0
2
4
6
8
10
12
14
16
18
Time (s)
Solution: In this example we need to use Equation 5 with g replaced by a. There are
four time intervals A, B, C and D. In each of these intervals we need to calculate the
acceleration of the runner.
Interval A: a = 4 m/s2
Interval B: a = 0
Interval C: a = - 2 m/s2
Interval D: a = 0
Distance covered in interval A:
1
1
y o + v o t + at 2 = 0 + 0 + × 4 × 2 2 = 8 m
2
2
Distance covered in interval B:
1
y o + v o t + at 2 = 0 + 8 × 8 + 0 = 64 m
2
Distance covered in interval C:
1
1
y o + v o t + at 2 = 0 + 8 × 2 − × 2 × 2 2 = 12 m
2
2
Distance covered in interval D
1
y o + v o t + at 2 = 0 + 4 × 4 + 0 = 16 m
2
Hence total distance traveled by the runner in 16 s is 8 + 64 +12 + 16 = 100 m
5
4. A student in Caracas did the free fall experiment in his Physics class and recorded
the data shown below. Find the value of the acceleration of gravity in Caracas by
plotting an appropriate graph.
y (m)
t (s)
0
10
20
30
40
50
60
70
80
90
100
0
1.41
2.00
2.45
2.83
3.16
3.46
3.74
4.00
4.24
4.47
Solution: We first create new table calculating t2 as shown. Then we plot a graph of y
versus t2 and find the slope. Multiply the slope by 2 to get the acceleration due to
gravity in Caracas.
t (s)
t2(s2)
0
10
20
30
40
50
60
70
80
90
100
0
1.41
2.00
2.45
2.83
3.16
3.46
3.74
4.00
4.24
4.47
0
1.99
4.00
6.00
8.01
9.98
11.97
13.99
16.00
17.98
19.98
120
100
80
y (cm)
y (m)
60
m = 5.0045 m/s2
40
20
0
0
5
10
15
20
t2 (s2)
The acceleration due to gravity is g = 2*m = 2 * 5.0045 = 10 m/s2
6
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