Respuesta :
Answer: 3.63 s
The car would accelerate due to force of friction. Initially the car would have to overcome static force of friction.
Static force of friction:
[tex]F_s=\mu_smg[/tex]
Where[tex]\mu_s[/tex] is the coefficient of static friction, mg is the weight.
Let m be the mass of the car and a be the acceleration, then
[tex]ma=\mu_smg \Rightarrow a=\mu_sg[/tex]
It is given that, [tex]\mu_s=1.00[/tex]
[tex]a=1.00\times 9.8 m/s^2= 9.8 m/s^2[/tex]
Now, using the equation of motion:
[tex] v-u=at[/tex] we can find the shortest time in which the car would accelerate from[tex]u= 0 \hspace{1mm} to \hspace{1mm} v=80 mph=80\frac{miles}{h} \times 1.6 \frac{km}{mile} \times \frac{5 m/s}{18km/h}=35.56 m/s[/tex]
[tex] \Rightarrow t=\frac{v-u}{a}=\frac{35.56 m/s-0}{9.8m/s^2}=3.63 s[/tex]
For a typical rubber-on-concrete friction, the shortest time in which a car could accelerate from 0 to 80 mph is 3.65 seconds.
Given the data in the question;
Since the car accelerated from 0 to 80.
Initial velocity; [tex]u = 0m/s[/tex]
Final velocity; [tex]v = 80mph = 35.7632 m/s[/tex]
To determine the shortest time
We use the expression from the Force of Static Friction:
[tex]f_s = u_sN[/tex]
[tex]f_s = u_smg[/tex]
Also, From Newton's Second Law of Motion:
[tex]F = ma[/tex]
Now, from the diagram ( free body )
[tex]F = f_s[/tex]
Hence,
[tex]ma = u_smg\\a = u_sg[/tex]
We know that acceleration due to gravity;[tex]g = 9.8m/s^2[/tex] and in the question, [tex]u_s = 1.00[/tex]
So we substitute into the equation
[tex]a = 1.00 \ *\ 9.8m/s^2\\a = 9.8m/s^2[/tex]
Next, from the First Equation of Motion;
[tex]v = u + at[/tex]
Where v is the final velocity, u is the initial velocity, a is the acceleration.
We make time "t", the subject of the formula
[tex]t = \frac{v-u}{a}[/tex]
We substitute in our value
[tex]t = \frac{35.7632m/s- 0m/s}{9.8m/s^2} \\\\t = \frac{35.7632m/s}{9.8m/s^2} \\\\t = 3.649s\\\\t = 3.65s[/tex]
Therefore, for a typical rubber-on-concrete friction, the shortest time in which a car could accelerate from 0 to 80 mph is 3.65 seconds.
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