To solve the problem, we'll follow these steps:
(a) Calculate the deceleration
The deceleration can be calculated using the formula that relates the force of friction to the weight of the car. The force of friction (\(F_f\)) acting on the car is given by:
\[ F_f = \mu \cdot N \]
Where:
- \(\mu\) is the coefficient of friction (0.8),
- \(N\) is the normal force, which equals the weight of the car when on a horizontal surface, given by \(N = mg\) (where \(m\) is the mass and \(g\) is the acceleration due to gravity, approximately \(9.81 , \text{m/s}^2\)).
Now we calculate the normal force:
\[ N = m \cdot g = 1200 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 11772 , \text{N} \]
Now calculate the frictional force:
\[ F_f = \mu \cdot N = 0.8 \cdot 11772 , \text{N} = 9417.6 , \text{N} \]
Using Newton's second law, \(F = ma\), we can find the deceleration (\(a\)):
\[ F_f = m \cdot a \implies a = \frac{F_f}{m} = \frac{9417.6 , \text{N}}{1200 , \text{kg}} \approx 7.848 , \text{m/s}^2 \]
The deceleration is approximately:
\[ a \approx 7.85 , \text{m/s}^2 \]
(b) Calculate the distance travelled by the car before it comes to rest
We can use the kinematic equation relating initial velocity, final velocity, acceleration, and distance:
\[ v^2 = u^2 + 2a s \]
Where:
- \(v\) is the final velocity (0 m/s, since the car comes to rest),
- \(u\) is the initial velocity (20 m/s),
- \(a\) is the deceleration (-7.85 m/s², we take it as negative since it's a deceleration),
- \(s\) is the distance travelled.
Rearranging the equation to find \(s\):
\[ 0 = (20 , \text{m/s})^2 + 2(-7.85 , \text{m/s}^2)s \] \[ 0 = 400 - 15.7s \] \[ 15.7s = 400 \] \[ s = \frac{400}{15.7} \approx 25.5 , \text{m} \]
The distance travelled before coming to rest is approximately:
\[ s \approx 25.5 , \text{m} \]
(c) Calculate the time taken for the car to come to rest
We can use another kinematic equation that relates initial velocity, final velocity, acceleration, and time:
\[ v = u + at \]
Rearranging for \(t\):
\[ t = \frac{v - u}{a} \]
Substituting in the known values:
\[ t = \frac{0 - 20 , \text{m/s}}{-7.85 , \text{m/s}^2} \] \[ t \approx \frac{-20}{-7.85} \approx 2.55 , \text{s} \]
The time taken for the car to come to rest is approximately:
\[ t \approx 2.55 , \text{s} \]
Summary of Results:
- Deceleration: \( \approx 7.85 , \text{m/s}^2 \)
- Distance travelled: \( \approx 25.5 , \text{m} \)
- Time taken: \( \approx 2.55 , \text{s} \)
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