Complete Question
An aluminum "12 gauge" wire measures a diameter of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field E in the wire varies over time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is recorded in seconds.
At time 5 seconds, I = 1.2 A.
We need to find the charge Q traveling through a cross-section of the conductor from time 0 to time 5 seconds.
Answer:
The charge is 
Explanation:
The question indicates that
The wire’s diameter is 
The radius of the wire is 
Aluminum's resistivity is 
The electric field variation is described as

The charge is effectively given by the equation

Where A is the area expressed as

Thus,

Therefore

By substituting values
![Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5Cint%5Climits%5E%7Bt%7D_%7B0%7D%20%7B%20%5B%200.0004t%5E2%20-%200.0001t%20%2B0.0004%5D%20%7D%20%5C%2C%20dt)
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%20t%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
The question states that t = 5 seconds
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%205%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
![Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004%285%29%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20%285%29%5E2%7D%7B2%7D%20%2B0.0004%285%29%5D%20%7D)

Answer:
The direction in which a vehicle accelerates aligns with its velocity direction. However, the force of acceleration works against the car's speed.
Explanation:
The car’s initial acceleration can be found using:
v = v₀ + a t
a = (v-v₀) t
which assumes the initial speed is zero (v₀ = 0 m/s).
a = v / t
a = 300 / t
The acceleration vector matches the direction of the vehicle's movement.
Upon hitting the wall, a force is exerted in the reverse direction to halt the car, thus this acceleration opposes the vehicle’s speed. However, the module should be much greater since the stopping distance is minimal.
We installed a motion detector at one end of the track and placed a cart on it. We then attached a motorized fan to the cart, allowing it to propel the cart down the track. This aligns with my expectations based on the velocity graph, where the slope represents acceleration, which remains consistent over time.