Emergency Physics Help. (1 Viewer)

[Twickel]

Determines the relationship between velocity and stopping distance linking this to the force applied on the passenger or pedestrian, quantitatively.

How can I do this, anyone have any formulas and please provide examples.

 

[Aerath]

S = (mu²)/2F

S = Stopping distance (m)
m = mass of car
F = for of friction
u = initial velocity

Therefore, Stopping distance is directly proportional to initial velocity squared.

 

[Continuum]

S = (mu²)/2F

S = Stopping distance (m)
m = mass of car
F = for of friction
u = initial velocity

Therefore, Stopping distance is directly proportional to initial velocity squared.

Yep yep.

It's just 0.5mv²=Fs rearranged. This place really should have like... a maths equation creator thingy.

 

[Twickel]

Ok so 2f= mu^2/s

So if initial velocity is doubled and stopping distance remains the same then force is 4 time greater. Correct?
s= mu^2/2F

plot some numbers in let me think come on Twick.

ok so if velocity increases and distance remains = force increases. If velocity remains constant and distance increases force decreases. Now what can I say about what happens to velocity in relation to increases decreases of distance and force.

 

[Twickel]

Is that the only formula that links these 3 things together?