Basic dynamics/acceleration question

[Mike_Stokes]

Homework Statement

A 1000kg car is traveling at 27.0 m/s. Determine the force required to stop it in 70.0 m.

The Attempt at a Solution

I attempted to treat this question as if I knew how many seconds it took to stop, and did this:

70.0m / 27.0m/s = 2.59259s

2.59259s / 27.0m/s = 0.096

1000kg / 0.096 = 10414.29N

F = 10.4kN

But then I realized that this is all predicated on the idea that we are moving at 27.0 m/s throughout the whole 70.0m rather than accelerating. Now I have no idea what to do, and both of my textbooks are proving useless. Please help, and sorry if the question is messed up or hard to decipher, this is my first post.

 

[PhanthomJay]

Homework Statement

A 1000kg car is traveling at 27.0 m/s. Determine the force required to stop it in 70.0 m.

The Attempt at a Solution

I attempted to treat this question as if I knew how many seconds it took to stop, and did this:

70.0m / 27.0m/s = 2.59259s

2.59259s / 27.0m/s = 0.096

1000kg / 0.096 = 10414.29N

F = 10.4kN

But then I realized that this is all predicated on the idea that we are moving at 27.0 m/s throughout the whole 70.0m rather than accelerating. Now I have no idea what to do, and both of my textbooks are proving useless. Please help, and sorry if the question is messed up or hard to decipher, this is my first post.

You should first understand and become familiar with the SUVAT equations of motion for constant acceleration which you can find in any text to solve for the acceleration, then correctly apply Newton's Laws.

 

[Simon Bridge]

"suvat" equations are sometimes called "kinematic equations"
you can also work stuff out from the velocity-time diagrams.

you should have had some coursework about force and acceleration
- if you have not seen v-t diagrams, then you have probably had some work on average velocity.

Using average velocity to do this will make most people here cringe, but some courses do it that way at first.

 

[Mike_Stokes]

"suvat" equations are sometimes called "kinematic equations"
you can also work stuff out from the velocity-time diagrams.

you should have had some coursework about force and acceleration
- if you have not seen v-t diagrams, then you have probably had some work on average velocity.

Using average velocity to do this will make most people here cringe, but some courses do it that way at first.

Ahh okay, yeah we've done kinematic equations, but I was having a hard time remembering them, and we never referred to SUVAT at all as an acronym. Thanks so much for the help guys, I finally got the right formula for the question:

a = (Vf^2 - Vi^2)/2D

F = 5.26kN

Thanks for helping my (now seeming stupid) question!

 

[HalfLostAndSearching]

I would approach this problem by first identifying the key variables and equations that are relevant to the situation. The key variables in this problem are the mass of the car (m), its initial velocity (v0), the distance it needs to stop (d), and the force required to stop it (F). The relevant equation for this situation is Newton's Second Law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration (F=ma).

Next, I would determine the acceleration of the car using the equation for average acceleration, which is the change in velocity (v) divided by the time (t) it takes for that change to occur (a=Δv/Δt). In this case, we know that the car starts at 27.0 m/s and comes to a stop, so the change in velocity (Δv) is -27.0 m/s (negative because the car is slowing down) and the time (Δt) is 2.59259s. Therefore, the average acceleration of the car is:

a = (-27.0 m/s) / (2.59259s) = -10.4 m/s^2

Note that the negative sign indicates that the car is decelerating or slowing down.

Finally, we can use Newton's Second Law to calculate the force required to stop the car:

F = ma = (1000 kg) (-10.4 m/s^2) = -10400 N

Again, the negative sign indicates that the force is in the opposite direction of the car's motion, which makes sense since we want to stop the car from moving forward.

In conclusion, the force required to stop the car in 70.0m is approximately 10400 N. It is important to note that this is the average force required over the entire distance of 70.0m, so the actual force may vary depending on how the car's brakes are applied and other factors.