How resistance affects car speed?

[markkoka]

Hello !
I have little problem understanding about how car rolling resistance , wind resistance and angular resistance affect car speed.


SPEED is km/h but resistance is in NEWTONS , i don't understand how that force affects car speed , like FORCE (N) converted to speed(km/h or mph)
How to find car real speed while before resistances it was 100 and after it will be slowed by resistance forces.


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About rolling resistance i know how to find it (i will get xxx NEWTONS ) but how it affects car speed?

I know that rolling resistance means that car wheel torque must be bigger than 100N to make car roll , but does that rolling resistance
also affect car final speed also? slows it down a little bit?

If speed without rolling resistance is 100 km/h or mph then does rolling resistance affect that speed ?
If yes how to find what will be car speed after i add to speed 100km/h or mph that rolling resistance.

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AIR RESISTANCE - i know how to find it , but how thes NEWTONS affects speed?
I got again let's say random number 200N is air resistance at speed 100km/h or mph
then how i find car real speed after thaT? That force must slow car down a little bit.


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ANGULAR (grade) resistance , basically its the resistance when car goes uphill and it will slow down by hill angle.
I know how to find that force , but again how it affects speed?

Lets say i have >>>>
weight 1400kg
and hill angle 5deg
it means 1400 * sin(5deg)=122kg

i just take this 122kg and add to car weight while i find out rolling resistance?
because car weight is bigger then rolling resistance become also bigger??
So it means rolling resistance still affects car final speed??


Sorry if the question looks dumb for you but i really need to understand logic behind this , its small part of big project what i have to do at school . Internet is full of info how to find out these forces but how i use it to reduce real speed is not explained :(

I would be thankful for little help .

 

[drvrm]

Sorry if the question looks dumb for you but i really need to understand logic behind this , its small part of big project what i have to do at school . Internet is full of info how to find out these forces but how i use it to reduce real speed is not explained :(

If one applies force on bodies it starts moving -so Newton force does change the state of rest or motion of a body -its by work energy relation.
so try to find work done by forces and it will come in joule -i.e. Newton-metre and this energy imparts kinetic energy of motion i.e. one half of mass times square of velocity - if the moving force is working a lot the mph velocity will be more'
similarly if opposing or resistance is there as forces opposing the motion ,they will do work in reducing the energy that is speed of the car/body.
one must learn how to calculate work done and relate it to kinetic energy to understand fully the role of forces acting on bodies.

 

[Delta2]

The missing link in your learning-understanding on how to relate force to speed is Newton's second Law

Force=mass x acceleration or $F=ma$ written also as $a=\frac{F}{m}$

and the definition of acceleration a
$a=\frac{v_f-v_i}{t}$
which holds when acceleration a (and also the force F) are constant with respect to time and space coordinates.

So for example if you have a force $F=-10N$ (the minus sign simply means that the force opposes the initial speed, simply put it is a resistance force ) that acts on a body of mass $m=1Kg$ and initial velocity $v_i=10m/s$ then u find from Newton's second law that $a=-10m/s^2$ and from the definition of acceleration equation you ll find that the final speed -AFTER the force has act for time t=1second on the body- is $v_f=v_i+at=10-10x1=0m/s$

 

[CWatters]

Markkoka .. Some of the forces you mention are constant and others like air resistance increase with speed. The engine also provides a force on the car. In short.. the car will accelerate until the net force on the car is zero. At that point the force generated by the engine (forwards) equals the sum of all the resistive forces (rearwards).