Physics, 01.08.2019 07:00 caitievincent9739

For typical rubber-on-concrete friction, what is the shortest time in which a car could accelerate from 0 to 50 mph? suppose that μs=1.00 and μk=0.80. express your answer to two significant figures and include the appropriate units.

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First, let's convert the speed from mph to m/s, keeping in mind that
1 mi = 1609 m
1 h = 3600 s
So
$v_f = 50 mph \cdot \frac{1609}{3600}=22.35 m/s$
The car starts with initial speed v_i=0

and reaches v_f

in a certain time $\Delta t$ .
In order for the car to accelerate, we must have that the force pushing the car (mass x acceleration) is at least equal to the frictional force, pointing into the opposite direction. We can consider only the dynamic frictional force, since the static frictional force acts only in the very first moment of the motion:
$ma = m g \mu_k$
where g=9.81 m/s^2

and $\mu_k=0.8$ .
The acceleration is also given by
$a= \frac{v_f-v_i}{\Delta t} = \frac{v_f}{\Delta t}$
since v_i=0

. Substituting in the first formula, we get
$\frac{v_f}{\Delta t}=g \mu_k$
From which we can find the time necessary to reach the mentioned velocity: $\Delta t= \frac{v_f}{g \mu_s}=2.85 s$

Answer from: notsosmart249

The shortest time in which the car could accelerate from 0 to 50 mph is 2.85 second.

Explanation:

Given information:

$\mu_s=1.00\\\mu_k=0.80$

Convert the velocity in $\text{m/s.}$

As:

$v_f=50\;mph\times (1609/3600) \\v_f=22.35\;\text{m/s}$

The car is having initial speed zero and reaches v_f in a certain time $\Delta t$

Now, considering only dynamic frictional force, since the static frictional force will act at very first moment of the motion:

$ma=mg\mu_k$

As, the acceleration is given by:

$a=\frac{v_f-v_i}{\Delta t}$ ; here, v_i=0

$v_f/\Delta t=g.\mu_k$

on putting the values in above equation:

$\Delta t=v_f/g\mu_k\\\Delts t=2.85\;s$

Hence, the shortest time in which the car could accelerate from 0 to 50 mph is 2.85 second.

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Answer from: Quest

Sorry can’t u just tryna get points

Answer from: Quest

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

so, refractive index of medium = 300000/200000 = 1.5

1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

∞ refractive index is usually for opaque mediums. so third medium is opaque medium.

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