Physics, 11.11.2019 07:31, YaBoiMando2061

Acar moving at 46 km/h skids 14 m with locked brakes. how far will the car skid with locked brakes at 92 km/h? assume that energy loss is due only to sliding friction. answer in units of m.

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answered: jholland03

at 92 km/h the car will skid 56 meters

Explanation:

Notice that the only force acting to stop the car is the force of friction which we assume is the same in both cases because we deal with the same car under similar pavement/tire conditions. Therefore the negative acceleration that stops the car must be the same in both cases. we don't know that acceleration but can use the letter a to represent its magnitude in our equations.

We will also use the subindex "1" to represent the physical quantities associated with the equations of motion for the first case (car moving at 46 km/h) such that $v_1, \, t_1, x_1$ represent the velocity, time to stop, and distance traveled under the negatively accelerated motion in this first case.

Similarly, we represent with subindex "2" the physical quantities associated with the second case (car moving at 92 km/h) and for which we need to find the distance covered under negatively accelerated motion.

Equations of motion for the first case (car breaking - negative acceleration) from an initial velocity v_1 to rest (zero velocity). Lets start with the equation for the acceleration: $-a=\frac{0-v_1}{t_1} \\-a=\frac{-v_1}{t_1}\\t_1=\frac{-v_1}{-a} \\t_1=\frac{v_1}{a}$

We use this expression for the stopping time in the other equation of motion that involves the distance x_1 covered in the stopping process:

$x_1=v_1*t_1-\frac{1}{2} a*t_1^2\\x_1=v_1(\frac{v_1}{a}) -\frac{1}{2} a*(\frac{v_1}{a}) ^2\\x_1=\frac{v_1^2}{a} -\frac{1}{2} \frac{v_1^2}{a} \\x_1=\frac{1}{2} \frac{v_1^2}{a}$

We can extract from here the expression for the magnitude of the acceleration that, since it is the same as for the second case, we can use in the new equations of motion associated with it:

$a=\frac{v_1^2}{2*x_1}$

For the second case we do a similar study considering that the car goes from velocity v_2 to rest, in a time t_2 , and covering a distance x_2 (what we need to find):

$-a=\frac{0-v_2}{t_2} \\-a=\frac{-v_2}{t_2}\\t_2=\frac{-v_2}{-a} \\t_2=\frac{v_2}{a}$

And in the equation for the distance covered:

$x_2=v_2*t_2-\frac{1}{2} a*t_2^2\\x_2=v_2(\frac{v_2}{a}) -\frac{1}{2} a*(\frac{v_2}{a}) ^2\\x_1=\frac{v_2^2}{a} -\frac{1}{2} \frac{v_2^2}{a} \\x_2=\frac{1}{2} \frac{v_2^2}{a}$

And we now use the expression for the magnitude of the acceleration from the first case (for which we know the distance covered) to replace it and find the new distance to full stop:

$x_2=\frac{1}{2} \frac{v_2^2}{a}\\x_2= \frac{1}{2}\frac{v_2^2\,2\,x_1}{v_1^2} \\x_2=\frac{v_2^2}{v_1^2} x_1$

We now just replace the values for the velocities and for the breaking distance for case 1. Notice that since we are dividing velocities that have the same units they cancel out and we don;t need to worry about converting units to the SI system. The answer for the breaking distance in case 2 will come in the same units as the breaking distance for case 1 (that is in meters). We proceed now to use the numerical values to get our

$x_2=\frac{v_2^2}{v_1^2} x_1\\x_2=\frac{92^2}{46^2} 14\\x_2=56$

Then, the car travelling at 92 km/h will skid 56 meters

answered: Guest

An atom (ours, not carroll's) consists of a positively charged nucleus, surrounded by one or more negatively charged particles called electrons. the positive charges equal the negative charges, so the atom has no overall charge; it is electrically neutral. most of an atom's mass is in its nucleus; the mass of an electron is only 1/1836 the mass of the lightest nucleus, that of hydrogen. although the nucleus is heavy, it is quite small compared with the overall size of an atom. the radius of a typical atom is around 0.1 to 0.25 nanometres (nm), whereas the radius of a nucleus is about 10-6 nm. * if an atom were enlarged to the size of the earth, its nucleus would be only 200 feet in diameter and could easily rest inside a small football stadium.read more on -

answered: Guest

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Acar moving at 46 km/h skids 14 m with locked brakes. how far will the car skid with locked brakes a...