Working With 1D Constant Acceleration Kinematics

In summary, the student is seeking help with a homework problem involving drawing a graph of velocity versus time for an object with constant acceleration. They are struggling with understanding how to show that vave=1/2(v0+v) and how to eliminate t in the equation v2=v02+2aΔx. They have completed the other two problems in the set and are familiar with the equations v-v0=∫vdt and vave=1/2(v0+v). They are also aware that they need two equations involving t to eliminate it.
  • #1
Entangled Cat
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Hello, this is my first post on PhysicsForums. I'm a first year student at the University of Kansas pursuing a Bachelor of Science in Physics and Astronomy (double majoring). The wording on my homework (for Honors General Physics 1) is a little bit strange to me so maybe some of you guys and girls can help me iron it out. Thanks in advance!

1. Homework Statement

Draw a graph of velocity versus time of an object starting with a velocity v0 and increasing speed with a constant acceleration(this is easy and not a problem). We know that v-v0=∫vdt (this is also quite obvious and easy to deduct), so the displacement is the area under the plot you just drew(simple integration). Show that for this case (constant acceleration): vave=1/2(v0+v) - (although I know this is true, I don't exactly know how to "show" it?), and then, by equating this result with the definition of vave and eliminating t(this is ultimately what hangs me up. I have no idea how to eliminate t. I know that the definition of vave) = Δx/Δt but I am not sure how to rid the equation of time), v2=v02+2aΔx

Homework Equations


v2=v02+2aΔx
vave) = Δx/Δt
ave=1/2(v0+v)

The Attempt at a Solution


Since this is more of a conceptual approach to understanding this idea, I have very little work done for this specific problem. I have finished the other two problems in the question set though (total of 3 problems).
 
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  • #2
Entangled Cat said:
v-v0=∫vdt
s-s0?
Entangled Cat said:
vave=1/2(v0+v)
This is just geometry. What shape have you drawn? How would you find its area?
Entangled Cat said:
and eliminating t
There seems to be something missing here in the instructions. To eliminate t you need two equations involving t. The other equation you need is the definition of acceleration.
 

FAQ: Working With 1D Constant Acceleration Kinematics

1. What is 1D constant acceleration kinematics?

1D constant acceleration kinematics is a branch of physics that studies the motion of objects in one dimension with a constant acceleration. It involves analyzing the position, velocity, and acceleration of an object over time.

2. How is acceleration related to velocity in 1D constant acceleration kinematics?

In 1D constant acceleration kinematics, acceleration is directly related to velocity. As acceleration increases, velocity also increases, and as acceleration decreases, velocity decreases. This relationship is described by the equation a = Δv/Δt, where a is acceleration, Δv is the change in velocity, and Δt is the change in time.

3. What is the difference between average acceleration and instantaneous acceleration?

Average acceleration is the overall change in velocity over a certain period of time, while instantaneous acceleration is the acceleration at a specific moment in time. Average acceleration can be calculated by dividing the change in velocity by the change in time, while instantaneous acceleration is determined by taking the derivative of the velocity-time graph at a given time.

4. How do you calculate displacement in 1D constant acceleration kinematics?

Displacement is the change in position of an object over time. In 1D constant acceleration kinematics, displacement can be calculated using the equation x = x0 + v0t + 1/2at^2, where x is the final position, x0 is the initial position, v0 is the initial velocity, t is the time, and a is the acceleration.

5. What is the difference between positive and negative acceleration in 1D constant acceleration kinematics?

Positive acceleration occurs when an object's velocity is increasing, while negative acceleration (also known as deceleration or retardation) occurs when an object's velocity is decreasing. Positive acceleration is represented by a positive value in calculations, while negative acceleration is represented by a negative value. A change in direction can also result in negative acceleration, even if the object's speed is increasing.

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