Question

It is friction that provides the force for a car to accelerate, so for high-performance cars the factor that limits acceleration isn't the engine; it's the tires. you may want to review ( pages 138 - 142) . part a for typical rubber-on-concrete friction, what is the shortest time in which a car could accelerate from 0 to 60 mph? suppose that μs=1.00 and μk=0.80.

Answer 1

Let the acceleration of the system is a and the time to reach 60 mph is t, so 

ax = (μs - μr) * g 
ax = (1.00-0.80) (9.8 m/s2) = 1.96 m/s2 
Then I used acceleration in this equation: 
vf = vi + ax * t 
60 mph = 0 mph + (1.96 m/s2) t 
t = (60 mph/1.96 m/s2)(0.447 m/s / 1 mph) 
t = 13.68 s

Answer 2

The shortest time in which a car could accelerate from 0 mph to 60 mph is $13.7s$.

Further explanation:

The opposite force acting on the body is known as frictional force. It always acts in the opposite direction of motion of body.

Concept used:

The force applied to a body to keep it at rest is known as the static friction force. It always acts opposite to the direction of motion of body. It is defined as the product of coefficient of friction and the normal force acting on the body.

The expression for the normal reaction of the body is given as.

$N = mg$  

The expression for the net force is given as.

${F_{net}} = ma$                                 …… (1)

The expression for the static friction is given as.

${F_s} = {\mu _s}N$

The expression for the balanced forces is given as.

${F_{net}} = {F_s} - {F_r}$

 

Substitute${\mu _s}N$ for ${F_s}$ and  for${F_r}$ in the above expression.

$\begin{aligned}{F_{net}}&={\mu _s}N-{\mu _r}N\\&= \left( {{\mu _s} - {\mu _r}} \right)N \\ \end{aligned}$

Substitute $mg$ for $N$ in above expression.

${F_{net}}=\left({{\mu _s}-{\mu _r}}\right)\left( {mg}\right)$       …… (2)

Compare equation (1) and (2) we get.

$a=g\left({{\mu _s}-{\mu _r}}\right)$                              …… (3)

Here, $a$ is the acceleration of the body, g is the acceleration due to gravity, ${\mu _s}$ is the coefficient of static friction and  is the coefficient of reactive force.

The expression for the first equation of motion is given as.

$v = u + at$      

                             

Rearrange the above expression for time is given as.

$\fbox{\begin\\t = \dfrac{{\left( {v - u} \right)}}{a}\end{minispace}}$                              …… (4)

Here, $v$ is the final velocity, $u$ is the initial velocity and $t$ is the time.

Substitute $1$ for ${\mu _s}$, $0.8$ for${\mu _r}$ and $9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}$ for $g$ in equation (3).

$\begin{aligned}a&=\left( {9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}}\right)\left( {1 - 0.8}\right)\\&=1.96\,{\text{m/}}{{\text{s}}^{\text{2}}} \\ \end{aligned}$

 

Substitute $0\,{\text{mph}}$ for $u$, $60\,{\text{mph}}$ for $v$ and $1.96\,{\text{m/}}{{\text{s}}^{\text{2}}}$ for $a$ in equation (4).

$\begin{aligned}t&=\frac{{\left( {60\,{\text{mph}}-0\,{\text{mph}}}\right)}}{{1.96\,{\text{m/}}{{\text{s}}^{\text{2}}}}}\\&=\frac{{60\,{\text{mph}}\left( {\frac{{0.447\,{\text{m/s}}}}{{1\,{\text{mph}}}}} \right)}}{{1.96\,{\text{m/}}{{\text{s}}^{\text{2}}}}}\\&= 13.7\,{\text{s}} \\ \end{aligned}$

 

Thus, the time required to accelerate a car is $\fbox{\begin\\13.7\, {\text{s}}\end{minispace}}$.

Learn more:

1.  Conservation of momentum https://brainly.com/question/9484203.

2.  Motion under friction https://brainly.com/question/7031524.

3. Net force on the body https://brainly.com/question/4033012.

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

Force, friction, Acceleration, acceleration due to gravity, normal, weight, mass, motion, sliding, sled, hill, inclined, plane, coefficient of friction, angle of inclination, 13.68 s, 13.69 s.