Respuesta :
Answer:b
Explanation:
Given
Initial velocity [tex]u=30 m/s[/tex]
Coefficient of kinetic friction [tex]\mu _k=0.6[/tex]
maximum deceleration [tex]a=\mu _kg=0.6\times 9.8=5.88 m/s^2[/tex]
Final velocity [tex]v=0[/tex]
using [tex]v=u+at[/tex]
[tex]0=30-5.88\times t[/tex]
[tex]t=\frac{30}{5.88}[/tex]
[tex]t=5.10 s[/tex]
Given the coefficient of kinetic friction and the initial velocity, the time taken for the car to come to a stop is approximately 5.10 seconds.
Hence, Option B) 5.10s is the correct answer.
Given the data in the question;
- Initial velocity; [tex]u = 30m/s[/tex]
Since the car was brought to a stop
- Final velocity; [tex]v = 0m/s[/tex]
- Coefficient of kinetic friction; [tex]u_k = 0.600[/tex]
Time taken for car to stop; [tex]t = \ ?[/tex]
First we determine the deceleration ( negative acceleration ) "a" due to friction:
[tex]a = -u_k*g\\\\a = -0.600 * 9.81m/s^2\\\\a = -5.88m/s^2[/tex]
Now, using 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 and t is the time taken.
We substitute our values into the equation
[tex]0m/s = 30.0m/s + (-5.88m/s^2 * t)\\\\(5.88m/s^2 * t) = 30m/s\\\\t = \frac{30m/s}{5.88m/s^2}\\\\t = 5.10s[/tex]
Given the coefficient of kinetic friction and the initial velocity, the time taken for the car to come to a stop is approximately 5.10 seconds.
Hence, Option B) 5.10s is the correct answer.
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