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 (\(\(u\)\)) = 7 m/s
- Final velocity (\(\(v\)\)) = 0 m/s (since the car comes to a stop)
- Stopping distance (\(\(s\)\)) = 10 m
- Mass of the car (\(\(m\)\)) = to be determined (but not needed to be specified for the final ratio)
2. Calculate the deceleration (\(\(a\)\)) using the kinematic equation:
The kinematic equation that relates initial velocity, final velocity, acceleration, and distance is:
\(\[
v^2 = u^2 + 2as
\]\)
Since \(\(v = 0\)\) (the car stops), we get:
\(\[
0 = u^2 + 2as
\]\)
Solving for the acceleration \(\(a\)\):
\(\[
a = -\frac{u^2}{2s}
\]\)
Plugging in the given values:
\(\[
a = -\frac{(7\ \text{m/s})^2}{2 \times 10\ \text{m}} = -\frac{49}{20}\ \text{m/s}^2 = -2.45\ \text{m/s}^2
\]\)
The negative sign indicates that this is a deceleration.
3. Calculate the braking force (\(\(F\)\)):
According to Newton's second law, force is given by:
\(\[
F = m \cdot a
\]\)
Substituting \(\(a = -2.45\ \text{m/s}^2 \)\):
\(\[
F = m \cdot (-2.45\ \text{m/s}^2) = -2.45m\ \text{N}
\]\)
The negative sign indicates that the force is acting in the opposite direction to the motion.
4. Calculate the weight (\(\(W\)\)) of the car:
The weight (\(\(W\)\)) of an object is given by:
\(\[
W = m \cdot g
\]\)
where \(\(g\)\) is the acceleration due to gravity (\(\(9.81\ \text{m/s}^2\)\)):
\(\[
W = m \cdot 9.81\ \text{m/s}^2 = 9.81m\ \text{N}
\]\)
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:
\(\[
\text{Ratio} = \frac{|F|}{W} = \frac{2.45m}{9.81m} = \frac{2.45}{9.81}
\]\)
Simplifying this ratio:
\(\[
\text{Ratio} = \frac{2.45}{9.81} \approx 0.25
\]\)
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.
