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What would the graph look like on a position verse time plot with constant acceleration?
If the constant acceleration is positive, the graph would be an exponential (x2) graph. If there is constant acceleration, then velocity is always increasing, making the position change at an ever increasing rate.
Why is the slope of a distance velocity squared graph straight and a distance velocity graph is not?
When acceleration is constant, one equation of kinematics is: (final velocity)^2 = 2(acceleration)(displacement) + (initial velocity)^2. When you are graphing this equation with displacement or position of the x-axis and (final velocity)^2 on the y-axis, the equation becomes: y = 2(acceleration)x + (initial velocity)^2. Since acceleration is constant, and there is only one initial velocity (so initial velocity is also constant), the equation becomes: y = constant*x + constant. This looks strangely like the equation of a line: y = mx + b. Therefore, the slope of a velocity squared - distance graph is constant, or there is a straight line. Now, when you graph a velocity - distance graph, the y axis is only velocity, not velocity squared. So if: v^2 = mx + b. Then: v = sqrt(mx + b). Or: y = sqrt(mx + b). This equation is not a straight line. For example, pretend m = 1 and b = 0. So the equation simplifies to: y = sqrt(x). Now, make a table of values and graph: x | y 1 | 1 4 | 2 9 | 3 etc. When you plot these points, the result is clearly NOT a straight line. Hope this helps!
Will the acceleration be zero if velocity is constant in velocity time graph?
Yes, by definition, Acceleration is the derivative of velocity. From Calculus, we know, the derivative of any constant function is simply 0. Therefore, any velocity time graph containing a constant velocity function will have an acceleration function = 0. c'=0 or c d/dx=0 also, there is a very simple, and nice progression to things in the field of Kinematics. where given a function, whether it be position, velocity, or acceleration, you can find the other through calculus. the progression for integration goes as follows, jolt (the feeling you get when acceleration changes), acceleration, velocity, position. And the reverse for derivatives. Having a velocity function of something like 55mph, think of it as driving on the freeway, absent of outside forces of friction, air resistance and hills, you never have to accelerate (positively, or negatively) to maintain speed, and the math backs that up. 55 is constant, and the derivative is 0. now if you had a velocity function such as 2x^2 + 5, also known as a parabolic function, you would take the derivative and get an acceleration of 4x, and a jolt of 4, and if you integrate to get position, you would have 2/3x^3+5x. Its all about the math.
Can something have an acceleration while moving at a constant velocity?
No. A object can have acceleration while moving at a constant speed (like the earth in its orbit around the sun). But once you use the word "velocity", you've expanded the decription of its motion to include the direction as well as the speed. The definition of acceleration is a change in either speed or direction, so if there is acceleration, then either the speed or the direction of the velocity (or both) is changing, so the velocity is not constant.
What does an object that is speeding up at a constant rate in the positive direction what does the velocity vs time graph look like?
A line angled upward
What would the graph look like on a position verse time plot with constant acceleration?
If the constant acceleration is positive, the graph would be an exponential (x2) graph. If there is constant acceleration, then velocity is always increasing, making the position change at an ever increasing rate.
What does the graph of acceleration vs time look like for something going a constant way?
Because acceleration is the derivative of velocity, you can determine what an acceleration vs. ... t graph are straight and horizontal, i.e. the object moves at a constant velocity, the slopes of those lines are 0 , and so the a vs. t graph should show a straight, horizontal line at y=0 (along the x -axis).
What does the graph of accelaration vs.time look like for something going a constant velocity?
Constant velocity implies zero acceleration, so you would have a horizontal line, identical to the x-axis.
What is the importance of uniform acceleration graph in physics what is instantaneous velocity?
Uniform acceleration graphs help visualize how an object's velocity changes over time. They show a constant rate of change in velocity, which can be used to calculate properties like displacement and time. Instantaneous velocity is the velocity of an object at a specific moment in time, representing the object's speed and direction at a given instant.
How is constant acceleration represented on a velocity time graph?
When an object moves in a straight line with constant acceleration, the equation describing its position (s) in terms of time (t) is a quadratic function like s = a t2 + b t + c, where a, b, and c are constants. The graph of such an equation is a parabola. However, if u plot velocity against time, the function is linear, and the graph is a straight line.
Why is the slope of a distance velocity squared graph straight and a distance velocity graph is not?
When acceleration is constant, one equation of kinematics is: (final velocity)^2 = 2(acceleration)(displacement) + (initial velocity)^2. When you are graphing this equation with displacement or position of the x-axis and (final velocity)^2 on the y-axis, the equation becomes: y = 2(acceleration)x + (initial velocity)^2. Since acceleration is constant, and there is only one initial velocity (so initial velocity is also constant), the equation becomes: y = constant*x + constant. This looks strangely like the equation of a line: y = mx + b. Therefore, the slope of a velocity squared - distance graph is constant, or there is a straight line. Now, when you graph a velocity - distance graph, the y axis is only velocity, not velocity squared. So if: v^2 = mx + b. Then: v = sqrt(mx + b). Or: y = sqrt(mx + b). This equation is not a straight line. For example, pretend m = 1 and b = 0. So the equation simplifies to: y = sqrt(x). Now, make a table of values and graph: x | y 1 | 1 4 | 2 9 | 3 etc. When you plot these points, the result is clearly NOT a straight line. Hope this helps!
Will the acceleration be zero if velocity is constant in velocity time graph?
Yes, by definition, Acceleration is the derivative of velocity. From Calculus, we know, the derivative of any constant function is simply 0. Therefore, any velocity time graph containing a constant velocity function will have an acceleration function = 0. c'=0 or c d/dx=0 also, there is a very simple, and nice progression to things in the field of Kinematics. where given a function, whether it be position, velocity, or acceleration, you can find the other through calculus. the progression for integration goes as follows, jolt (the feeling you get when acceleration changes), acceleration, velocity, position. And the reverse for derivatives. Having a velocity function of something like 55mph, think of it as driving on the freeway, absent of outside forces of friction, air resistance and hills, you never have to accelerate (positively, or negatively) to maintain speed, and the math backs that up. 55 is constant, and the derivative is 0. now if you had a velocity function such as 2x^2 + 5, also known as a parabolic function, you would take the derivative and get an acceleration of 4x, and a jolt of 4, and if you integrate to get position, you would have 2/3x^3+5x. Its all about the math.
How is deceleration represented on a velocity per time graph?
It is radial the velocity in a direction towards or away from a fixed point of reference (the origin) at a given time. The velocity time graph takes no account of motion in a direction across the radial direction.
What does an object that is speeding up at a constant rate in the positive direction what does the velocity vs time graph look like?
A line angled upward
Can something have an acceleration while moving at a constant velocity?
No. A object can have acceleration while moving at a constant speed (like the earth in its orbit around the sun). But once you use the word "velocity", you've expanded the decription of its motion to include the direction as well as the speed. The definition of acceleration is a change in either speed or direction, so if there is acceleration, then either the speed or the direction of the velocity (or both) is changing, so the velocity is not constant.
For an object that is speeding up at a constant rate in the positive direction what does the velocity vs time graph look like?
A line angled upward
How do you find the initial velocity from an acceleration -time graph?
You cannot find the initial velocity from an acceleration-time graph on its own. What you can do is find the change in velocity over the first T units of time. This is the area under the a-t graph from t = 0 to t = T. This may be a simple calculation of the area of a rectangle or a trapezium or may require integration. But, like all integrations, you end up with an unknown constant - in this case the initial velocity. You must have some additional information - usually a boundary condition - that allows you to find this constant and so the initial velocity.
