Answered

A motor car of mass running at the rate of 7 m/s can be stopped by its
brakes in 10 m. Prove that the total resistance to the car's motion, when
the brakes are on, is one quarter of the weight of the car.
[3]



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.