How do I calculate the braking distance for a car on dry road?

Question

The car had been traveling at 40 mph when its driver applied the brakes. What is the distance needed?

The coefficient for dry road from my experiment is .642
I think i'm suppose to use the equation

d= -vi^2/2a

but when does the use of the coefficient come in?

How would I do this if the problem does give me a weight?

I meant DOES NOT give me a MASS

Answer 1

easy ;-)

the friction force, call it F, is equal to the normal force (the force perpendicular to a surface - in this case, the car's weight), times the friction coefficient, call it k. And the weight would be equal to the mass, m, times g. So you'd have

F = k * m * g

now the acceleration is the force divided by the mass, so:

a = F / m = k*g = (0.642)*(9.81) (in metric)

the formula you suggest to use is the correct one ;-)

now i'm used to metric so i'll convert your mph to m/s, 40mph is 17.8 m/s.

so you have

d = (17.8)^2 / 2*(0.642)*(9.81)
d = 25 metres

if you want feet, and don't want metric in between, then you'd have to use 32 ft/s2 and not 9.81 m/s2 for g. And you'd have to convert your 40mph in ft/sec (about 58.3 ft/s).

hope this helps

Answer 2

i think you should do an energy balance. when the driver applies the brakes, the car will slow down and stop due to friction. Without friction, it will continue running.

mv^2/2 = Fd where v is the velocity, F is the frictional force and d is the distance.

F = (coefficient)(ma) where m is mass, a is acceleration due to gravity.. so, ma is the weight of the car

substituting F to the first equation,

mv^2/2 = (coefficient)(ma)d

dividing both sides by m,

v^2/2 = (coefficient)(a)(d)
therefore, d = v^2/(2a(coefficient))

we have similar equation but you only lack the (coefficient)... the negative sign is conventional. since you have a negative sign, the "a" you should use should be negative for "d" to be positive. However, in my equation, a positive "a" should be used for "d" to be positive.

Answer 3

Stopping sight distance is the sum of the break reaction distance (the distance travelled between the time the driver sees an obstruction to when the brakes are applied) and the braking distance (the distance travelled while braking the vehicle to a stop).

The equation y=.6x2+1.1x can be used to find this, where x is equal to miles per hour and y is the final Stopping Sight Distance. (The original equation listed here was incorrect)

Answer 4

Forget the equation, conserve energy:

=>Change in kinetic energy=work done by friction
Since sliding Friction=(fric. coeff)mg

=>1/2mV^2=(fric. coeff)mg X d.......where d=breaking dist.

=>d=(V^2)/{2(fric. coeff)g}

1 ml =1.609 km
=>40 mlph=64.36 kmph=17.88 metreps

=>d=(17.88)^2/(2*0.642*9.8)
=>d=25.41 metre