physics study of motion
Regents Physics
Constant Velocity/ Acceleration Lab 10-3-13
Problem:
Graphical Analysis of Constant Velocity and Accelerated Motion.
Theory:
Gravitational acceleration is constant on Earth g=9.8m/s2 Therefore, when the golf ball is dropped, the acceleration will be equal to gravitational acceleration agb=9.8m/s2 Given there is no air resistance, this means that when the golf ball is dropped from a given distance, according to the formulas, the golf ball will accelerate at 9.8m/s2 until reaching the floor.
Using graphs of the data collected, information on the acceleration and velocity of both the golf ball and cart can be shown. From these initial position/ time graphs, using slope can calculate the velocity of the object moving, and even the acceleration. To calculate these movements from the initial graphs, the slope formula shown below is used.
mp/t=change in d/change in t
(where: m=slope, d=distance, t=time)
So for initial position/ time graphs: mp/t=velocity Using the information collected from these formulas, other graphs can be created. With new information, velocity/ time graphs can be created, and using a similar formula, for the one used to calculate the velocity, acceleration can be calculated.
mv/t= change in v/change in t
(where: v=velocity)
So for the calculated velocity/ time graphs: mv/t= acceleration (a)
For a constant velocity graph, the acceleration of that graph should be equal to zero per any given amount of time, while the constant acceleration graph should be equal to a calculated number (mv/t) per any given amount of time.
A constant velocity distance/ time graph should have a constant increasing positive slope, while the velocity/ time graph should have a slope of zero at the y value of the calculated slope of the initial graph. For the constant acceleration graphs, the distance/ time graphs should have an exponentially increasing positive slope, the
Constant Velocity/ Acceleration Lab 10-3-13
Problem:
Graphical Analysis of Constant Velocity and Accelerated Motion.
Theory:
Gravitational acceleration is constant on Earth g=9.8m/s2 Therefore, when the golf ball is dropped, the acceleration will be equal to gravitational acceleration agb=9.8m/s2 Given there is no air resistance, this means that when the golf ball is dropped from a given distance, according to the formulas, the golf ball will accelerate at 9.8m/s2 until reaching the floor.
Using graphs of the data collected, information on the acceleration and velocity of both the golf ball and cart can be shown. From these initial position/ time graphs, using slope can calculate the velocity of the object moving, and even the acceleration. To calculate these movements from the initial graphs, the slope formula shown below is used.
mp/t=change in d/change in t
(where: m=slope, d=distance, t=time)
So for initial position/ time graphs: mp/t=velocity Using the information collected from these formulas, other graphs can be created. With new information, velocity/ time graphs can be created, and using a similar formula, for the one used to calculate the velocity, acceleration can be calculated.
mv/t= change in v/change in t
(where: v=velocity)
So for the calculated velocity/ time graphs: mv/t= acceleration (a)
For a constant velocity graph, the acceleration of that graph should be equal to zero per any given amount of time, while the constant acceleration graph should be equal to a calculated number (mv/t) per any given amount of time.
A constant velocity distance/ time graph should have a constant increasing positive slope, while the velocity/ time graph should have a slope of zero at the y value of the calculated slope of the initial graph. For the constant acceleration graphs, the distance/ time graphs should have an exponentially increasing positive slope, the