What Is the Minimum Deceleration Needed to Avoid an Accident?

[pedro]

help needed with accelleration problem...urgent!

I am having trouble wiht one of the questions on my first physics HW asignment.
I've tried solving it different ways but can't seem to get the correct answer.

the question reads:
An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When the engineer first sees the car, the locomotive is 280m from the crossing, and its speed is 24m/s.
if the engineer's reaction time is .55s, what should be the magnitude of the minimum deceleration to avoid an accident?

I have been using the formula:
Vfinal^2 - Vinitial^2 = 2a(xfinal - xinitial)

V being velocity, a being accelleration and x being position.

I solve this equation for a and begin substituding in.
I know that my Vinitial is 280 - 34(.55) because you have to take out ground covered in the engineers reaction time.

but when I substitute in everything else I alwasy get -1.079
because of the wording of the question I have tried it with and without the negative sign, but it is still wrong.

any help would be greatly appreciated
pedro

 

[dextercioby]

The sign should be minus,because it's a DECELERATION.The modulus is much,much smaller than what u've gotten.Since the logics and the formula u used are good (24*0.55,not 34*0.55,okay??),i advide you to check the arithmetics...

Daniel.

 

[wais]

Hello Pedro,

I understand your urgency and will try my best to help you with your acceleration problem. Firstly, it is great that you have identified the formula to use for this problem. However, it seems like you may have made a mistake in your calculations.

Let's break down the problem together. We know that the initial velocity (Vinitial) is 24m/s and the final velocity (Vfinal) is 0m/s (since the train needs to come to a complete stop to avoid an accident). We also know that the distance (x) between the train and the car is 280m. The reaction time (t) is given as 0.55s.

Now, using the formula Vfinal^2 - Vinitial^2 = 2a(xfinal - xinitial), we can substitute in the values we know:

(0)^2 - (24)^2 = 2a(280 - 280 - (24*0.55))

Simplifying this, we get:

-576 = -26.4a

Dividing both sides by -26.4, we get:

a = 21.82m/s^2

This is the magnitude of the minimum deceleration needed to avoid an accident. Note that it is a positive value, meaning that the train needs to decelerate in the positive direction (opposite to its initial direction of motion).

I hope this helps you with your problem. If you still have trouble, don't hesitate to reach out for further assistance. Good luck with your assignment!