Respuesta :
Answer:
The car was 12.8m/s fast when the driver applied the brakes.
Explanation:
The equations of motion of the car in the horizontal and vertical axes are:
[tex]x: f_k=ma\\\\y: N-mg=0[/tex]
Since the kinetic friction is defined as [tex]f_k=\mu_kN[/tex] and [tex]N=mg[/tex] we have:
[tex]\mu_kmg=ma\\\\a=\mu_kg[/tex]
Next, from the kinematics equation of speed in terms of distance, we have:
[tex]v^2=v_0^2-2ax\\\\v^2=v_0^2-2\mu_kgx[/tex]
Since the car came to a stop, the final velocity [tex]v[/tex] is zero, and we get:
[tex]0=v_0^2-2\mu_kgx\\\\v_0=\sqrt{2\mu_kgx}[/tex]
Finally, plugging in the known values, we obtain:
[tex]v_0=\sqrt{2(0.300)(9.81m/s^2)(28.0m)}\\\\v_0=12.8m/s[/tex]
It means that the car was 12.8m/s fast when the driver applied the brakes.
The car was moving at the speed of 12.8 m/s, when the driver applied the brakes.
From kinamatic equation,
[tex]\bold {v^2 = v_0^2-2ax}\\[/tex]
Since, car stops the final veocity will be zero. and
Acceleration [tex]\bold {a = \mu_kg}[/tex]
So,
[tex]v_0 = \sqrt {2\mu k_gx}[/tex]
Where,
Vo - initial speed =?
[tex]\mu[/tex] - friction constant = 0.3
g - gravitational acceleration = 9.8 m/s²
x - distance = 25 m
Put the values,
[tex]v_0 = \sqrt {2\times 0.3 \times 9.81 \times 0.25 m}\\\\v_0 = 12.8\ m/s[/tex]
Therefore, the car was moving at the speed of 12.8 m/s, when the driver applied the brakes.
To know more about kinamatic equation,
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