Acar moving with constant acceleration covers the distance between two points 60 m apart in 6.0 s. its speed as it passes the second point is 15 m/s. (a) calculate the speed at the first point. (b) calculate the acceleration. (c) at what prior distance from the first point was the car at rest?

The speed in the first point is: 4.98m/s

The acceleration is: 1.67m/s^2

The prior distance from the first point is: 7.42m

Explanation:

For part a and b:

We have a system with two equations and two variables.

We have these data:

X = distance = 60m

t = time = 6.0s

Sf = Final speed = 15m/s

And We need to find:

So = Inicial speed

a = aceleration

We are going to use these equation:

We are going to put our data:

With these equation, you can decide a method for solve. In this case, We are going to use an egualiazation method.

If we analyze the situation, we need to have an aceleretarion  greater than cero. We are going to choose a = 1.67m/s^2

After, we are going to determine the speed in the first point:

For part c:

We are going to use:

answer: the answer is a

electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. shielding is typically applied to enclosures to isolate electrical devices from their surroundings, and to cables to isolate wires from the environment through which the cable runs. electromagnetic shielding that blocks radio frequency electromagnetic radiation is also known as rf shielding.

the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia