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Suppose a car accelerates constantly from rest. The work done is $W=Fd$, so the net force times the distance traveled is:

$$ ma \cdot \frac{1}{2}a t^2 = m \cdot \Delta v/t \cdot \frac{1}{2} \cdot \Delta v/t \cdot t^2 = \frac{1}{2} mv^2 ~~~~~~(\mathrm{final~velocity}) $$

Is this energy the car has, gained from its whole traject or is this the energy it has while it is moving with its final velocity? And what if is it is not accelerating from rest, but from a certain velocity to a final velocity, should the v squared in the formula be the difference between the initial and final velocity?

Btw, I had watched this easy derivation of the formula, which led me to the above questions:

https://www.youtube.com/watch?v=zt0ov2C5svw

I know there are more questions related on my topic, but there is a lot of university physics involved rather than just some easy algebra.

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  • $\begingroup$ Please use MathJax for you equations. $\endgroup$ – Bob D Feb 2 at 17:47
  • $\begingroup$ Are you familiar with the work-energy theorem? $\endgroup$ – Bob D Feb 2 at 17:49
  • $\begingroup$ Yes, I know about kinetic energy, work and potential energy. $\endgroup$ – mathomato Feb 2 at 17:50
  • $\begingroup$ I was trying to format the question but missed spacing dollar signs ,sorry $\endgroup$ – Aditya Prakash Feb 2 at 18:09
  • $\begingroup$ No problem, thanks for your answers. $\endgroup$ – mathomato Feb 2 at 18:10
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Is this energy the car has, gained from its whole trajectory or is this the energy it has while it is moving with its final velocity?

At each point in the trajectory the car has a certain velocity and thus a certain amount of kinetic energy. How much energy it has at a given point depends on the velocity it has attained at the point.

And what if is it is not accelerating from rest, but from a certain velocity to a final velocity

The only difference here is the car already has kinetic energy based on its initial velocity before being accelerated. But the change in kinetic energy will depend on the acceleration and time or distance.

should the v squared in the formula be the difference between the initial and final velocity?

The applicable work-energy equation is

$$Fd=\frac{mv_{f}^2}{2}-\frac{mv_{i}^2}{2}$$

Hope this helps.

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  • $\begingroup$ Great, thank you. $\endgroup$ – mathomato Feb 2 at 18:09
  • $\begingroup$ So the answer is acceptable to you? $\endgroup$ – Bob D Feb 2 at 18:10
  • $\begingroup$ Yeah, in my dutch high school book they just give you the formulas without a lot of explanation, this is surely acceptable. $\endgroup$ – mathomato Feb 2 at 18:11
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    $\begingroup$ You are aware there is an "accept" button next to each submitted answer? Lots of new contributors are not. Look at all the answers and pick the one you like best. Or, if none completely answers your question wait until more answers are submitted. $\endgroup$ – Bob D Feb 2 at 18:17
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The formula $$W = Fd$$ is only true for a force that does not change over a distance or an average force over a distance. Let's derive kinetic energy in a general way using some calculus.

Suppose we apply some force F over an infintesimal displacement dx. The work done will be $$dW = Fdx$$

To find the total work done over a larger displacement, we can integrate. $$W = \int_{x_i}^{x_f} F dx$$

But $F = ma$. Making this substitution: $$W = \int_{x_i}^{x_f} (ma) dx$$

What is acceleration? It is the derivative of velocity with respect to time. $$a = \frac{dv}{dt}$$ Now put this into our work expression. $$W = \int_{x_i}^{x_f} (m\frac{dv}{dt})dx$$

Rearrange the fraction to get

$$W = \int_{x_i}^{x_f} (m\frac{dx}{dt})dv$$

But $\frac{dx}{dt} = velocity$! Make this substitution and change the limits of integration to initial and final velocity because we are now integrating with respect to velocity:

$$W = \int_{v_i}^{v_f} (mv)dv$$

Evaluate this integral! With a constant mass, you will get the result

$$W = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = \frac{1}{2}m(v_f^2 - v_i^2)$$

We see now that all that matters is the change in velocity!

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  • $\begingroup$ "The formula 𝑊=𝐹𝑑 is only true for a force that does not change." That is not correct. $F$ in the formula can be the average force applied over the distance $d$. In fact the average value is used in applying the work-energy theorem to crashes to determine the average impact force. $\endgroup$ – Bob D Feb 2 at 20:54
  • $\begingroup$ Oh, you are right! I edited my answer. Thank you $\endgroup$ – pwatts Feb 3 at 5:32
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Whether the body started from rest or it started from a certain initial velocity ,does not matter.You can set t=0 whenever you want to and the Kinetic energy will be dependent only on the instantaneous velocity for a point mass when measured in the same frame of reference.
The formula $$K.E.= \frac{1}{2} mv^2$$ will give you the energy acquired by the body by virtue of its velocity at that instant ,the paths to that velocity could be different ,you could accelerate uniformly or non-uniformly and come to that particular velocity and you will have the same kinetic energy.
When you set t=0 at a certain initial velocity $v$ ,you are basically adding $F.d$ to the already present $\frac{1}{2}mv^2$ so there is no question of kinetic energy being the difference of final and initial kinetic energies ,so you can safely say that the formula gives you the energy required to bring a body to a certain velocity from rest.

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$$W=\int \vec F. d\vec s=\int m\frac{d\vec v}{dt}.d\vec s=m\int \vec v.d\vec v=\frac{1}{2}m(v_f^2-v_i^2)$$ (assuming m is constant)

In order to state the work-energy theorem, physicist had defined the term $\frac{1}{2}mv^2$ as Kinetic Energy.

Kinetic energy can be defined capacity of 'moving thing' to do work. This means that the moving thing if stops by some external force how much work it could do. Clearly, it is $\frac{1}{2}mv^2$

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  • $\begingroup$ Good point specifying constant mass from a mathematical point of view. But it should be noted that Newton's second law applies only to constant mass objects. $\endgroup$ – garyp Feb 2 at 18:22

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