Calculating the Stopping distance

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    Stopping distance
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The discussion centers on calculating the propulsive power and minimum braking distance for a car under specific conditions. The propulsive power delivered by the engine while cruising at 120 km/hr is calculated to be 15 kW using the formula P = 450 * (120/3.6). For the braking distance from 100 km/hr, participants clarify that the mass of the car is not necessary for the calculation, as the relevant factors are the coefficient of friction, gravitational acceleration, and initial/final speeds. The equation D = V^2 / (2 * μ * g) is confirmed as appropriate for determining stopping distance. The conversation emphasizes the importance of understanding the role of mass in dynamics versus the parameters needed for specific calculations.
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Homework Statement


A car of mass 1000 kg is cruising at 120 km/hr. At this velocity the drag and friction forces that the engine needs to work against is equivalent to 450 N.

(i) What is the propulsive power being delivered by the engine when cruising? (4 marks)

(ii) From a velocity of 100 km/hr, the driver suddenly sees a traffic jam ahead and performs an emergency braking, resulting in locking the four wheels. Calculate the minimum braking distance required to bring the car to a full stop.

Assume the following: The coefficient of friction between the tyres and road is μ = 0.8.
Consider the weight of the vehicle to be equally distributed on each wheel while braking The drag or other friction forces are not significant in this case. g = 9.8 m/s2

Homework Equations


D=V/2*0.8*9.81

The Attempt at a Solution


All i know for the equation is D=V/2*0.8*9.81. However i am not sure how to implement the mass into the equation, Or if there is a different one to use.
 
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The equation you quote is fine for part two. Why do you wish to bring mass into it?
For part one, you need some more standard equations. What ones do you know about power, force, mass, acceleration?
 
I worked out Q 1 = 15KW
By using the P= 450*(120/3.6)=15KW

But for part be i assumed the weight must be added to the equation because it states the mass of the car. Or is that figure irrelevant to work out the distance ?
 
I also forgot to add this to the question..
The driver cannot avoid hitting the vehicle at the back of the queue. At the moment of impact, the driver's car has a remaining velocity of 15 km/hr and hits a stationary van that has a mass 1400 kg.
Assume that both vehicles are in neutral and not braking at the moment of impact. Neglect the effects of friction.

Calculate the velocities of each vehicle after impact if the coefficient of restitution of the vehicles' bumpers is 0.85.
 
xavierengineering said:
I also forgot to add this to the question..
The driver cannot avoid hitting the vehicle at the back of the queue. At the moment of impact, the driver's car has a remaining velocity of 15 km/hr and hits a stationary van that has a mass 1400 kg.
Assume that both vehicles are in neutral and not braking at the moment of impact. Neglect the effects of friction.

Calculate the velocities of each vehicle after impact if the coefficient of restitution of the vehicles' bumpers is 0.85.
That's the part where you need the mass of the car.
 
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I believe you need the mass of the car to get the drag force on it while skidding to a stop.
 
insightful said:
I believe you need the mass of the car to get the drag force on it while skidding to a stop.
No. We are given g, the coefficient of friction, and the initial and final speeds, and we wish to find a distance. None of those involves a mass dimension. Consequently (by dimensional analysis) being told one mass as well cannot be of use.
 
haruspex said:
No. We are given g, the coefficient of friction, and the initial and final speeds, and we wish to find a distance. None of those involves a mass dimension. Consequently (by dimensional analysis) being told one mass as well cannot be of use.
Oops, of course, you're right. My only defense is that the engineer in me is more comfortable going the "F=ma" route.
 
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