Respuesta :
Let the time taken be t
Initial speed (u) = 0 m/s
Final speed (v) = 60 mph = 26.82 m/s
If the static friction is acting on the car then it will accelerate more.
Acceleration due to static friction (a) = μg = 1 × 9.8 = 9.8 m/s^2
Now, using the first equation of motion to find time:
v = u + at
[tex]t = \frac{v-u}{a}[/tex]
[tex]t = \frac{26.82 - 0}{9.8}[/tex]
t = 2.74 seconds
Hence, the time taken by the car to reach 60 mph from 0 mph is 2.74 seconds.
The shortest time in which a car could accelerate from [tex]0[/tex] to [tex]60[/tex] mph is [tex]\fbox{2.737 s}[/tex].
Further Explanation:
Newton states three laws on motion.
1. The first law expressed as “If a body is in motion remains in motion and if in rest remains in rest until or unless an external force is applied on it.”
2. The second law expressed as “The force applied on the object is equal to the change in the momentum of the object per change in the time.”
3. The third law expressed as “If a force applied on the object then equal and opposite force will be applied by the object.”
Friction force is the force that always applied on the moving object in the opposite direction of the motion, in other words it opposes the motion of the object. It depends on the coefficient of the friction and the normal reaction of the object.
The value of the coefficient of the friction depends on the nature of the surface.
Given:
The initial speed is [tex]0 mph[/tex].
The final speed is [tex]60 mph[/tex].
The value of the coefficient of the static friction is [tex]1.00[/tex].
The value of the coefficient of the kinetic friction is [tex]0.80[/tex].
Concept:
The acceleration is inversely proportional to the time therefore to calculate the shortest time acceleration should be maximized.
The acceleration is maximum when the coefficient of the friction is maximum therefore we should take coefficient of static friction as it has high value than the coefficient of kinetic friction.
Apply Newton’s 2nd law of motion.
The net force required to move the car is equal to the friction force i.e.
[tex]ma = {\mu _s}N[/tex] …… (1)
Here, [tex]a[/tex] for acceleration, [tex]m[/tex] is the mass of the object, [tex]{\mu _s}[/tex] is the coefficient of friction and [tex]N[/tex] is the normal reaction.
The normal reaction is equal to the force due to gravity.
Therefore, [tex]N=mg[/tex]
Substitute [tex]mg[/tex] for [tex]N[/tex] in equation (1).
[tex]\begin{aligned}ma&={\mu _s}\left({mg}\right)\\ma&={\mu _s}mg\\a&={\mu _s}g\\\end{aligned}[/tex]
Substitute [tex]1.00[/tex] for [tex]{\mu _s}[/tex] and [tex]9.8{\text{ m/}}{{\text{s}}^2}[/tex] for [tex]g[/tex] in the above expression.
[tex]\begin{aligned}a&=\left( {1.00} \right)\left( {9.8{\text{ m/}}{{\text{s}}^2}} \right)\\&=9.8{\text{ m/}}{{\text{s}}^2}\\\end{aligned}[/tex]
The expression for the acceleration is:
[tex]\fbox{\begin\\a=\dfrac{{v - u}}{t}\end{minispace}}[/tex]
Here, [tex]v[/tex] is the final velocity, [tex]u[/tex] is the initial velocity and [tex]t[/tex] is the time.
Substitute [tex]0 mph[/tex] for [tex]u[/tex], [tex]60 mph[/tex] for [tex]v[/tex] and [tex]9.8{\text{ m/}}{{\text{s}}^2}[/tex] for [tex]a[/tex] in the above equation.
[tex]\begin{aligned}9.8{\text{ m/}}{{\text{s}}^2}&=\frac{{\left( {60 - 0} \right){\text{ mph}}}}{t}\\t&=\frac{{60{\text{ mph}}\left( {\frac{{1609.34{\text{ m}}}}{{1{\text{ mile}}}}} \right)\left( {\frac{{1{\text{ h}}}}{{3600{\text{ s}}}}} \right)}}{{9.8{\text{ m/}}{{\text{s}}^2}}}\\&=2.737{\text{ s}}\\\end{aligned}[/tex]
Therefore, the shortest time in which a car could accelerate from [tex]0[/tex] to [tex]60 mph[/tex] is [tex]\fbox{2.737 s}[/tex].
Learn more:
1. Find the magnitude of the Friction. https://brainly.com/question/8367599
2. The acceleration of the box. https://brainly.com/question/7031524
3. Find the approximate value of the friction force on the box. https://brainly.com/question/10132341
Answer Details:
Grade: Collage
Subject: Physics
Chapter: Kinematics
Keywords:
Typical. Rubber-on-concrete, shortest, time, car accelerate, friction, motion, oppose, 0 mph, 60 mph, {\mu _s} = 1.00, {\mu _k} = 0.80, 2.737 s.
