Question

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.

Answer 1

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 2

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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