For an automobile with wheel base 2.5m, the centre of gravity lies 0.6 m above the ground and 1.15 m in front of the rear axle. The automobile is moving on level ground with a speed of 45 km/hr. Find the minimum stopping distance when
(a) Only rear wheels are braked
(b) Only front wheels are braked
(c) Brakes are applied to all the wheels
Take the value of coefficient of friction equal to 0.5.
To find the minimum stopping distance, we need to calculate the deceleration of the automobile.
First, let's convert the speed from km/hr to m/s:
Speed = 45 km/hr = (45 * 1000) / 3600 = 12.5 m/s
Now, let's calculate the deceleration for each case:
(a) Only rear wheels are braked:
In this case, all the weight of the automobile is assumed to act at the rear axle. The deceleration can be calculated using the formula:
Acceleration = (Coefficient of friction) * (Gravity)
Where:
Coefficient of friction = 0.5
Gravity = 9.81 m/s^2
Acceleration = 0.5 * 9.81 = 4.905 m/s^2
Using the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s as the automobile is coming to a stop)
u = initial velocity (12.5 m/s)
a = acceleration (4.905 m/s^2)
s = stopping distance (unknown)
0 = (12.5)^2 + 2 * (-4.905) * s
Solving this equation will give us the stopping distance.
(b) Only front wheels are braked:
In this case, all the weight of the automobile is assumed to act at the front axle. The deceleration can be calculated using the same formula as in case (a):
Acceleration = 0.5 * 9.81 = 4.905 m/s^2
Using the equation of motion:
0 = (12.5)^2 + 2 * (-4.905) * s
The stopping distance will be the same as in case (a).
(c) Brakes are applied to all the wheels:
In this case, the weight distribution between the front and rear axles needs to be considered. Let's assume that the weight is evenly distributed 50% at the front and 50% at the rear.
The deceleration can be calculated using the formula:
Acceleration = (Coefficient of friction) * (Gravity) * (Weight distribution at axle)
Where:
Weight distribution at axle = (Weight of automobile * Distance of center of gravity from axle) / (Wheelbase)
Weight of automobile = Weight distribution at axle + Weight distribution at axle = 1 (kg) (assuming the automobile weighs 1 kg for simplicity)
Weight distribution at axle = 0.5
Distance of center of gravity from rear axle = 1.15 m
Acceleration = 0.5 * 9.81 * (0.5 * 1.15 / 2.5) = 1.12085 m/s^2
Using the equation of motion:
0 = (12.5)^2 + 2 * (-1.12085) * s
Solving this equation will give us the stopping distance.
Please note that the stopping distance calculated here is the distance traveled while braking until the automobile comes to a complete stop.
To find the minimum stopping distance for each scenario, we need to consider the factors that affect the stopping distance of a vehicle - the initial speed, the deceleration, and the time it takes to come to a stop.
The deceleration of a vehicle can be calculated using the equation:
deceleration = friction coefficient * acceleration due to gravity
where acceleration due to gravity is approximately 9.8 m/s^2.
Now let's calculate the stopping distances for each scenario:
(a) Only rear wheels are braked:
In this case, the deceleration will act only on the rear wheels, causing the vehicle to rotate around the rear axle. The center of gravity will exert a moment about the rear wheels, resulting in a pitching motion.
To calculate the deceleration, we can use the equation:
deceleration = friction coefficient * acceleration due to gravity
= 0.5 * 9.8
= 4.9 m/s^2
The stopping distance can be calculated using the equation:
stopping distance = v^2 / (2 * deceleration)
where v is the initial velocity.
Converting the speed from km/hr to m/s:
initial velocity = 45 km/hr = (45 * 1000) / (60 * 60) m/s
stopping distance = (v^2) / (2 * deceleration)
= ((45 * 1000) / (60 * 60))^2 / (2 * 4.9)
≈ 98.04 m
Therefore, the minimum stopping distance when only rear wheels are braked is approximately 98.04 m.
(b) Only front wheels are braked:
In this case, the deceleration will act only on the front wheels, causing the vehicle to rotate around the front axle. The center of gravity will exert a moment about the front wheels, resulting in a pitching motion.
The deceleration remains the same:
deceleration = friction coefficient * acceleration due to gravity
= 0.5 * 9.8
= 4.9 m/s^2
Using the same equation as before to calculate the stopping distance:
stopping distance = v^2 / (2 * deceleration)
= ((45 * 1000) / (60 * 60))^2 / (2 * 4.9)
≈ 98.04 m
Therefore, the minimum stopping distance when only front wheels are braked is also approximately 98.04 m.
(c) Brakes are applied to all the wheels:
In this case, the deceleration will act on all the wheels, providing a balanced braking force to the vehicle and minimizing pitching motion.
The deceleration remains the same:
deceleration = friction coefficient * acceleration due to gravity
= 0.5 * 9.8
= 4.9 m/s^2
Using the same equation as before to calculate the stopping distance:
stopping distance = v^2 / (2 * deceleration)
= ((45 * 1000) / (60 * 60))^2 / (2 * 4.9)
≈ 98.04 m
Therefore, the minimum stopping distance when brakes are applied to all the wheels is also approximately 98.04 m.
In conclusion, the minimum stopping distance for all scenarios (a), (b), and (c) is approximately 98.04 meters.