Stopping distance and time of a vehicle

• Lucaburra
In summary: The positive root is the time you want.In summary, the coefficient of friction and initial velocity are used to determine the distance and time it takes for a car to stop with its brakes applied. The equation used is dist = V0*t + .5*a*t^2, where a<0 or -4.71 m/sec^2. The time taken for the car to stop is dependent on the initial velocity and coefficient of friction, and is found by taking the positive root of the equation. Mass does not affect the time taken for the car to stop.
Lucaburra

Homework Statement

The coefficient of friction is .48. The brakes are applied to a 1.05 x10^3 car traveling at 95km/hr. How far does the car move before it stops ? How long does it take for the car to stop?

Homework Equations

Fg = Fn, Fa= u x Fn, a=Fa/m, d=Vf^2-Vi^2/2a, t=d/v

The Attempt at a Solution

Fg = Fn =-9.81m/s^2 x 1050 kg = 10300 N
Fa = u x Fn = .48 x 10300 N = 4944.2 N
a = F/m = 4944.2 N / 1050 Kg = 4.71 m/s
d = (Vf^2 - Vi^2)/2a =
95 km/hr converts to 26.4 m/s
so if Vf = 0 then -26.4 ^2 = 697.0 m/s^2
697.0 m/s^2 / 2 x 4.71 m/s = 74.0 m
then t = d/v so time is 74m/ 26.4 m/s = 2.80 secs

Im just not convinced I have done this correct...?

You have computed the distance correctly but your time is wrong. You should divide by the average velocity or use:

dist = V0*t + .5*a*t^2, where a<0

t=v/a
ma=mgμ

t=v/gμ
Thus time taken for a body to stop depends only on initial velocity, coefficient of friction.
Mass is not in the equation.

Going back to my post, the average velocity is half the initial velocity which doubles your answer for the time. The time also is the root of:

dist = V0*t + .5*a*t^2, where a<0 or -4.71 m/sec^2 which you computed above.

Your calculations seem to be correct. It is always important to double check your work and make sure your units are consistent throughout the problem. In this case, you correctly converted the initial velocity from km/hr to m/s, but in your final calculation for time, you used meters instead of kilometers. Therefore, the final answer for time should be 2.80 seconds, not 0.00280 seconds. Other than that, your solution is correct and follows the appropriate equations for calculating stopping distance and time. Good job!

What is stopping distance and time?

Stopping distance and time refers to the distance a vehicle travels and the time it takes to come to a complete stop after the brakes have been applied.

What factors affect stopping distance and time?

Stopping distance and time can be affected by factors such as the speed of the vehicle, the condition of the road surface, the condition of the tires and brakes, and the weight and size of the vehicle.

How is stopping distance and time calculated?

The stopping distance and time of a vehicle can be calculated using the formula: stopping distance = initial velocity² / (2 x deceleration), where initial velocity is the speed of the vehicle and deceleration is the negative acceleration due to the brakes being applied.

Why is it important to understand stopping distance and time?

Understanding stopping distance and time is important for ensuring safe driving practices. It allows drivers to anticipate the distance they need to stop their vehicle and adjust their speed accordingly, especially in emergency situations.

How can stopping distance and time be reduced?

Stopping distance and time can be reduced by maintaining a safe driving speed, regularly checking and maintaining the condition of the vehicle's brakes and tires, and keeping a safe distance from other vehicles on the road.

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