Answer :
To solve this problem, we'll need to use the principles of kinematics and dynamics to prove the given statement. Here is a step-by-step solution:
1. Identify the given quantities:
- Initial velocity ([tex]\(u\)[/tex]) = 7 m/s
- Final velocity ([tex]\(v\)[/tex]) = 0 m/s (since the car comes to a stop)
- Stopping distance ([tex]\(s\)[/tex]) = 10 m
- Mass of the car ([tex]\(m\)[/tex]) = to be determined (but not needed to be specified for the final ratio)
2. Calculate the deceleration ([tex]\(a\)[/tex]) using the kinematic equation:
The kinematic equation that relates initial velocity, final velocity, acceleration, and distance is:
[tex]\[ v^2 = u^2 + 2as \][/tex]
Since [tex]\(v = 0\)[/tex] (the car stops), we get:
[tex]\[ 0 = u^2 + 2as \][/tex]
Solving for the acceleration [tex]\(a\)[/tex]:
[tex]\[ a = -\frac{u^2}{2s} \][/tex]
Plugging in the given values:
[tex]\[ a = -\frac{(7\ \text{m/s})^2}{2 \times 10\ \text{m}} = -\frac{49}{20}\ \text{m/s}^2 = -2.45\ \text{m/s}^2 \][/tex]
The negative sign indicates that this is a deceleration.
3. Calculate the braking force ([tex]\(F\)[/tex]):
According to Newton's second law, force is given by:
[tex]\[ F = m \cdot a \][/tex]
Substituting [tex]\(a = -2.45\ \text{m/s}^2 \)[/tex]:
[tex]\[ F = m \cdot (-2.45\ \text{m/s}^2) = -2.45m\ \text{N} \][/tex]
The negative sign indicates that the force is acting in the opposite direction to the motion.
4. Calculate the weight ([tex]\(W\)[/tex]) of the car:
The weight ([tex]\(W\)[/tex]) of an object is given by:
[tex]\[ W = m \cdot g \][/tex]
where [tex]\(g\)[/tex] is the acceleration due to gravity ([tex]\(9.81\ \text{m/s}^2\)[/tex]):
[tex]\[ W = m \cdot 9.81\ \text{m/s}^2 = 9.81m\ \text{N} \][/tex]
5. Calculate the ratio of the braking force to the weight of the car:
To find the ratio, we’ll take the magnitude of the braking force and compare it to the weight:
[tex]\[ \text{Ratio} = \frac{|F|}{W} = \frac{2.45m}{9.81m} = \frac{2.45}{9.81} \][/tex]
Simplifying this ratio:
[tex]\[ \text{Ratio} = \frac{2.45}{9.81} \approx 0.25 \][/tex]
Thus, we have proved that the total resistance to the car's motion, when the brakes are on, is one quarter of the weight of the car.
1. Identify the given quantities:
- Initial velocity ([tex]\(u\)[/tex]) = 7 m/s
- Final velocity ([tex]\(v\)[/tex]) = 0 m/s (since the car comes to a stop)
- Stopping distance ([tex]\(s\)[/tex]) = 10 m
- Mass of the car ([tex]\(m\)[/tex]) = to be determined (but not needed to be specified for the final ratio)
2. Calculate the deceleration ([tex]\(a\)[/tex]) using the kinematic equation:
The kinematic equation that relates initial velocity, final velocity, acceleration, and distance is:
[tex]\[ v^2 = u^2 + 2as \][/tex]
Since [tex]\(v = 0\)[/tex] (the car stops), we get:
[tex]\[ 0 = u^2 + 2as \][/tex]
Solving for the acceleration [tex]\(a\)[/tex]:
[tex]\[ a = -\frac{u^2}{2s} \][/tex]
Plugging in the given values:
[tex]\[ a = -\frac{(7\ \text{m/s})^2}{2 \times 10\ \text{m}} = -\frac{49}{20}\ \text{m/s}^2 = -2.45\ \text{m/s}^2 \][/tex]
The negative sign indicates that this is a deceleration.
3. Calculate the braking force ([tex]\(F\)[/tex]):
According to Newton's second law, force is given by:
[tex]\[ F = m \cdot a \][/tex]
Substituting [tex]\(a = -2.45\ \text{m/s}^2 \)[/tex]:
[tex]\[ F = m \cdot (-2.45\ \text{m/s}^2) = -2.45m\ \text{N} \][/tex]
The negative sign indicates that the force is acting in the opposite direction to the motion.
4. Calculate the weight ([tex]\(W\)[/tex]) of the car:
The weight ([tex]\(W\)[/tex]) of an object is given by:
[tex]\[ W = m \cdot g \][/tex]
where [tex]\(g\)[/tex] is the acceleration due to gravity ([tex]\(9.81\ \text{m/s}^2\)[/tex]):
[tex]\[ W = m \cdot 9.81\ \text{m/s}^2 = 9.81m\ \text{N} \][/tex]
5. Calculate the ratio of the braking force to the weight of the car:
To find the ratio, we’ll take the magnitude of the braking force and compare it to the weight:
[tex]\[ \text{Ratio} = \frac{|F|}{W} = \frac{2.45m}{9.81m} = \frac{2.45}{9.81} \][/tex]
Simplifying this ratio:
[tex]\[ \text{Ratio} = \frac{2.45}{9.81} \approx 0.25 \][/tex]
Thus, we have proved that the total resistance to the car's motion, when the brakes are on, is one quarter of the weight of the car.
