Incomplete question as there is so much information is missing.The complete question is here
A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?
Answer:
Distance traveled=240 m
Explanation:
Given data
Initial velocity of car v₀=0 m/s
Final velocity of car vf=24 m/s
Distance traveled by car S=120 m
To find
Distance does the traffic travel
Solution
To find the distance first we need to find time, for time first we need acceleration
So
![(V_{f})^{2}=(V_{o})^{2}+2aS\\ So\\a=\frac{(V_{f})^{2}-(V_{o})^{2} }{2S}\\ a=\frac{(24m/s)^{2}-(0m/s)^{2} }{2(120)}\\a=2.4 m/s^{2}](https://tex.z-dn.net/?f=%28V_%7Bf%7D%29%5E%7B2%7D%3D%28V_%7Bo%7D%29%5E%7B2%7D%2B2aS%5C%5C%20%20So%5C%5Ca%3D%5Cfrac%7B%28V_%7Bf%7D%29%5E%7B2%7D-%28V_%7Bo%7D%29%5E%7B2%7D%20%7D%7B2S%7D%5C%5C%20a%3D%5Cfrac%7B%2824m%2Fs%29%5E%7B2%7D-%280m%2Fs%29%5E%7B2%7D%20%7D%7B2%28120%29%7D%5C%5Ca%3D2.4%20m%2Fs%5E%7B2%7D)
As we find acceleration.Now we need to find time
So
![V_{f}=V_{i}+at\\t=\frac{V_{f}-V_{i}}{a}\\t=\frac{(24m/s)-(0m/s)}{(2.4m/s^{2} )}\\t=10s](https://tex.z-dn.net/?f=V_%7Bf%7D%3DV_%7Bi%7D%2Bat%5C%5Ct%3D%5Cfrac%7BV_%7Bf%7D-V_%7Bi%7D%7D%7Ba%7D%5C%5Ct%3D%5Cfrac%7B%2824m%2Fs%29-%280m%2Fs%29%7D%7B%282.4m%2Fs%5E%7B2%7D%20%29%7D%5C%5Ct%3D10s)
Now for distance
So
![Distance=velocity*time\\Distance=(24m/s)*(10s)\\Distance=240m](https://tex.z-dn.net/?f=Distance%3Dvelocity%2Atime%5C%5CDistance%3D%2824m%2Fs%29%2A%2810s%29%5C%5CDistance%3D240m)
15x 9.8 x .3 = 44.1
15 x 9.8 x 1 = 147
147 - 44.1 = 102.9J
102.9 J
The answer is the last choice.
Its electrical potential energy stays the same because it has the same electric potential. The reason why is that moving the charge towards X does not change the distance of the negative charge between the plates. The Electrical potential energy of a particle is the result energy by virtue of its position from the electrical fields produce by the plates both positive and negative. Since the charge is still equidistant to each other (assuming based from the diagram) no change in terms of electrical energy consumption or work was done.
It accelerates in speed and also in change in direction.