what is the mathematical relationship b/w the velocity,thinking dist. and the braking distance of a car?

Question

explain fully to earn maximum point!

Answer 1

we can find it in this way:
Fs=1/2mv square

if F is a steady breaking force applied to a car of mass m moving initially with speed v, then s id breaking distance needed to stop it. Since F and m is constant we can say

"s is direcly propotional to square of speed"

The thinking distance is the distance travelled while the driver is reacting be4 applying the breaks)

hence, the stopping distance wud be


stopping distance=thinking distance+breaking distance

Answer 2

Good question.

The velocity relationship is v = u - at; where u is the initial velocity of the car, v is the final velocity = 0, a is the deceleration rate during the actual braking time t.

The deceleration is F/m = a; where F = kN and k = rolling friction coefficient and N = W = mg the normal weight of the car with mass m. We use rolling friction (pumping the brakes) because it is higher than static friction (locking the brakes).

Response to stimulus time S = T + R; where T is your thinking time and R is the reactiion time that follows thinking. The "stimulus" might be, for example, a child's ball rolling into the street or a stop light changing to red. These times are of course highly dependent on the scenario and the mental/physical capabilities of the driver. But normally, neither would be more than a second.

Thus, total time from stimulus to stop is TT = T + R + t(u,a); where t(u,a) = u/a = u//kN/m = um/kN = um/kmg = u/kg.

Then total stopping distance D = (T + R)u + 1/2 at^2 = (T + R)u + 1/2 kN/m (u/kg)^2 = (T + R)u + 1/2 (kmg/m) u^2/(kg)^2 = (T + R)u + 1/2 u^2/kg = (T + R)u + 1/2 ut Thus...

D = u(T + R) + ut/2 = u(T + R) + u^2/2kg; where t is the actual time of deceleration due to braking and u/2 is the average velocity over that time (u + v)/2 = u/2 because terminal velocity is v = 0. So if you mean the total distance, including the time to think and react, then D is the relationship. If you mean the distance while braking, then d = ut/2 or u^2/2kg

The u^2/2kg term is noteworthy. It clearly shows that actual braking distance (d) is proportional to the square of the initial velocity (u) and inversely proportional to the coefficient of rolling friction (k). This means that doubling the car's speed will quadruple the stopping distance d. It also means that, if it rains for example, and k goes down (the road gets slippery), the stopping distance d will also increase.

The (T + R)u term is also interesting. It shows that if the driver is distracted and thinking time is extended, D will be extended. For example, if the driver is talking on her cell phone and sees that ball rolling out into the street, it may take longer to think (T) about what to do. Or if the driver has his lap top on his knee while driving, it might take longer to react (R)because the computer is in the way.

Answer 3

1) assume during braking you have constant deceleration....

2) split this into motion before braking and motion during braking.

3) human reaction times = 300 milliseconds = .3 seconds

prior to applying breaks your car travels during your reaction time....

V = d/t', d = Vx t'
where d = distance, V = velocity, t' = reaction time

after applying breaks

Vf^2 = V^2 + 2ad

where Vf = final velocity = 0, V = initial velocity (same velocity as in previous part), a = rate of deceleration (varies by car), d = distance traveled during deceleration

so

0 = V^2 + 2ad

d = - V^2/2a

so total d = Vx t' - V^2/2a

so if you convert t' to hours (.3 s = 8.3 x 10^-5, express velocity in mph and assume deceleration = say -30 ft/s^2 ≈ -7.4 x 10^4 m/h^2

then....

d = 8.3 x 10^-5 x V + V^2/(1.48 x 10^5) in miles

if you multiply through by 5280 ft/mile

then d in feet is....

d = .44 V +.0359 V^2

example 60 mph, d =156 ft

Answer 4

According to the driver manual from the department of motor vehicles in California the thining time is 3/4th of a second. The decelleration of a car during the breaking period is the coefficient of friction between the tire and the road multiplies by the acceleration of gravity. The coefficient of friction for most cars are 1.5.The time it takes to come to a stop after reaction time is the time it takes to go from original velocity to zero. This forma an equation of ;

Orioginal velocity *(- COF *9.81m/s/s)*Time = 0

With that time you find solve the folowing;

Deceleration distance = thinking time X original velocity
+1/2(COF *9.81m/s/s)*time^2