To solve the problem, we can use the principles of physics related to motion and friction.
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Determine the deceleration caused by friction:
The force of kinetic friction (\( F_k \)) can be calculated using the equation:
\[ F_k = \mu_k \cdot N \]
where:
- \( \mu_k = 0.750 \) (the coefficient of kinetic friction),
- \( N = mg \) is the normal force (\( m \) is the mass of the car and \( g = 9.81 , \text{m/s}^2 \) is the acceleration due to gravity).
Therefore, the frictional force is:
\[ F_k = \mu_k \cdot mg = 0.750 \cdot mg \]
The deceleration (\( a \)) of the car can be found using Newton's second law (\( F = ma \)), where the only horizontal force acting on the car when it brakes is the frictional force, which acts opposite to the direction of motion:
\[ ma = -F_k \implies a = -\mu_k g \]
Substituting the known values:
\[ a = -0.750 \cdot 9.81 \approx -7.36 , \text{m/s}^2 \]
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Calculate the distance skidded using the kinematic equation:
We can use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance:
\[ v^2 = u^2 + 2as \]
where:
- \( v = 0 , \text{m/s} \) (final velocity),
- \( u = 20.0 , \text{m/s} \) (initial velocity when braking),
- \( a = -7.36 , \text{m/s}^2 \) (deceleration),
- \( s \) is the distance skidded.
Rearranging the equation to solve for \( s \):
\[ 0 = (20.0)^2 + 2(-7.36)s \]
\[ 0 = 400 - 14.72s \]
\[ 14.72s = 400 \implies s = \frac{400}{14.72} \approx 27.17 , \text{m} \]
So, for the initial speed of 20.0 m/s, the car skids approximately 27.17 meters before coming to a stop.
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Repeat the calculation for an initial speed of 40.0 m/s:
Using the same kinematic equation with \( u = 40.0 , \text{m/s} \):
\[ 0 = (40.0)^2 + 2(-7.36)s \]
\[ 0 = 1600 - 14.72s \]
\[ 14.72s = 1600 \implies s = \frac{1600}{14.72} \approx 108.92 , \text{m} \]
Therefore, if the car had been moving at 40.0 m/s, it would have skidded approximately 108.92 meters before stopping.
Summary:
- Distance skidded at 20.0 m/s: 27.17 meters
- Distance skidded at 40.0 m/s: 108.92 meters
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