How Does Newton's Second Law Apply to Highway Safety?

[oMovements]

Homework Statement
The minimum safe distance between vehicles on a highway is the distance a vehicle can travel in 2.0s at a constant speed. assume that a 1.2x10^3 kg car is traveling 72km/h when the truck ahead crashes into a northbound truck and stops suddenly.
a) if the car is at the required safe distance behind the truck, what is the separation distance?
b) If the average net braking force exerted by the car is 6.4x10^3 N [N], how long would it take the car to stop?
c) Determine whether a collision would occur. Assume that the driver's reaction time is an excellent 0.09s.

The attempt at a solution
3a) d = (vi+vf/2) t
= (1200+0/2) (2)
= 1200 m

b) Fnet = m*a
-6.4x10^3 = (1.2x10^2)a
a = -5.3 m/s^2

a = vf-vi/t
-5.3 = 0-1200/t
t = 226.4s

c) d = vit + 1/2at^2
d = (1200)(0.09) + 1/2 (-5.33)(0.09)^2
d = 108m
Which is the distance traveled in reaction

d = vit + 1/2at^2
= (1200)(225) + 1/2(-5.33)(225)^2
= 2.7x10^5 - 134915
= 135084m m

d = 13504 + 108
= 125192 m
so a collision would occur

Is this correct?

 

[Mentz114]

The first question, as I understand it, just requires the distance the car will go at 72Km/h in 2 seconds which is 2 times 72/3600. i.e. 40m. Your answer of 1200 m ( about a mile) is obviously wrong.

Part b),

a = f/m

v = v₀-at -> when v=0 t=v0/a = 72m/f.

 

[Rawrr!]

In 32a) to find the distance, you seemed to have used the mass of the car, in place of its velocity. That being said, i think your equation could be modified. The question is relating to the distance between the car and truck while in motion. the velocity remains constant.

distance = (velocity)x(time),

*make sure to convert the components into correct units. (km/h) -> (km/s) (since time is given in seconds)* *you will get a distance in terms of (km) when you do this, which you can then convert to (m) if you feel so compelled*

in b) When dealing with forces, it always helps to draw a force body diagram, letting you visualize the situation and see which direction the forces are going. This always helps me in determining the correct signs as well.

You have a force in the left direction, (by convenience, placing the car's velocity along the positive x-axis), meaning your stopping force is going to be negative.

-F_net = m_car x (-)a

This should make sense as the car is slowing down, the force is in the negative direction, and the car is accelerating to the left.

While you have the correct signs, you seemed to have neglected to convert the (km/h) of the velocity, into the (km/s) needed, since your time component is in seconds. It always helps to think about you answers in terms of the situation. With the time you have written down in your answer, it means it will take the car 226.4 seconds to come to a stop. (Which means it will take you over 3 minutes to come to a stop!) So i would always beg you to make sure you answer makes sense to the situation.

in 3c) you seemed to have done the kinematic equations correctly in this and the equations make sense in that the driver will travel a distance during reacting, and will also travel a distance during the braking, but again avoided to convert the (km/h) of the velocity, into (km/s). In order for the units to cancel, the time parameters must match. *Write down your units as well as the value into the equations, to make sure they cancel correctly, and that you finish with what you need! * I hope this helps!p.s. - Welcome to PF!

 

[oMovements]

In 32a) to find the distance, you seemed to have used the mass of the car, in place of its velocity. That being said, i think your equation could be modified. The question is relating to the distance between the car and truck while in motion. the velocity remains constant.

distance = (velocity)x(time),

*make sure to convert the components into correct units. (km/h) -> (km/s) (since time is given in seconds)* *you will get a distance in terms of (km) when you do this, which you can then convert to (m) if you feel so compelled*

in b) When dealing with forces, it always helps to draw a force body diagram, letting you visualize the situation and see which direction the forces are going. This always helps me in determining the correct signs as well.

You have a force in the left direction, (by convenience, placing the car's velocity along the positive x-axis), meaning your stopping force is going to be negative.

-F_net = m_car x (-)a

This should make sense as the car is slowing down, the force is in the negative direction, and the car is accelerating to the left.

While you have the correct signs, you seemed to have neglected to convert the (km/h) of the velocity, into the (km/s) needed, since your time component is in seconds. It always helps to think about you answers in terms of the situation. With the time you have written down in your answer, it means it will take the car 226.4 seconds to come to a stop. (Which means it will take you over 3 minutes to come to a stop!) So i would always beg you to make sure you answer makes sense to the situation.

in 3c) you seemed to have done the kinematic equations correctly in this and the equations make sense in that the driver will travel a distance during reacting, and will also travel a distance during the braking, but again avoided to convert the (km/h) of the velocity, into (km/s). In order for the units to cancel, the time parameters must match.


*Write down your units as well as the value into the equations, to make sure they cancel correctly, and that you finish with what you need! * I hope this helps!


p.s. - Welcome to PF!

Thanks, worked out the question and the answer makes sense. :)

 

[Nickie]

I cannot validate the accuracy of the calculations without knowing the units used for the given values. However, the approach and equations used appear to be correct. It is important to note that the minimum safe distance may vary depending on the speed and weight of the vehicles involved, as well as external factors such as road conditions and driver reaction time. It is always best to maintain a safe distance and drive with caution to prevent collisions.