[ REPOST FOR REFERENCE ] A car picking up speed at a uniform rate from rest passes three VECO posts in succession. The posts are 91 meters apart along the road. The car takes 10 seconds from the first to the second post, and 6 seconds from the second to the third post. Calculate:

the acceleration of the car
the distance from the starting point of the car to the first post
the car’s velocity at each post.

Question

[ REPOST FOR REFERENCE ] A car picking up speed at a uniform rate from rest passes three VECO posts in succession. The posts are 91 meters apart along the road. The car takes 10 seconds from the first to the second post, and 6 seconds from the second to the third post. Calculate:

the acceleration of the car
the distance from the starting point of the car to the first post
the car’s velocity at each post.

kittybasil · Answer

This is a remake of another post due to it being posted twice and everything being crazy in the responses. ALL OF THIS IS MY OWN WORK in response. Original posts are HERE and here Reposting in progress...

kittybasil · Answer

kittybasil wrote:

I will answer here instead of the other one to avoid mess Firstly let us establish what we need to find 1. car's acceleration* 2. distance from car's starting point (t=0) to first post 3. car's velocity at post 1 and post 2 (each) *I have a question, is the acceleration over the entire scenario or a specific part of this situation?

This is in response to the original poster because the first part is unclear. Anyway, moving on to actual work:

First let's calculate for part two, "distance from the starting point of the car to the first post" let $d$ be the distance from the starting point to the first post, $t$ be the time it takes to reach the first post from rest, and $a$ be the magnitude of the uniform acceleration. $$d=\frac{1}{2}a_1t^2$$$$d+91=\frac{1}{2}a_2(t+10)^2$$$$d+182=\frac{1}{2}a_3(t+16)^2$$ We should solve the system for $d$ in each one.

Note: the second equation's time is $t$ plus 16 because 10+6=16 seconds passed since bypassing the first post, and the distance is $d$ plus 182 because two posts have been bypassed with 91 meters in between each. And 91+91=182

But what about acceleration $a$? Well, let us view acceleration as the following formula:$$a=\frac{distance}{time\cdot time}=\frac{d}{t^2}$$Yes, you square the time value, because acceleration is velocity over time and velocity is distance over time. See the below image for an example explanation of how we calculate acceleration.

kittybasil · Answer

(The acceleration variables have subscripts because they aren't referring to the same situation, HOWEVER acceleration should be a constant rate so they should all be equivalent.) Technically, acceleration and velocity involve direction, but the car is literally on a flat plane (the road) and going in a singular forward direction, so we don't need to worry about that.

Anyway, we are solving for variable $d$ as the distance from rest (t=0, again) to the first post. This is actually a system of equations involving 3 equations and 3 variables, so there are multiple ways to solve this: 1. Gaussian elimination (aka matrices) 2. substitution 3. Cramer's rule

Since explaining the two fancy-sounding ones will take a LOT of time and effort and end up making this post long-winded and confusing, I have instead searched up the steps on Symbolab for substitution. HERE is the link (click the blue text) to those steps. ~~I can also type them up but I need a break from spending 5 hours trying to figure out a 3x3 system of equations by hand without knowing if I was 100% correct~~

Based on this, $d=\frac{4459}{250}=17.836\text{ meters}$, $t=7\text{ seconds}$ and $a=\frac{91}{120}= 0.758\overline{3}\approx0.7583\text{ m/s}^2$. You now have the answer for part three (distance from start point to the first post) AND part one (acceleration of the car)! Remember, $a=a_2=a_3$, all acceleration values are the same constant rate of increasing velocity.

kittybasil · Answer

Note: Please don't reply even if I'm not on the question, I might be taking a break from solving but be assured I am actively working on it in my free time (Since this is an extensive project I will be periodically taking time off in between solving parts of the question.)

I will be back to work on part two later!

kittybasil · Answer

That's gonna be a long "later" tbh. Classes just threw a gajillion amount of work at me 😔

kittybasil · Answer

If anyone wants to contribute feel free, but if you are not solving the question do not answer. Also please don't correct me unless you plan on redoing the entire question (in which case please do not do it on my own post)... I put a lot of work into making this.

AnimeWeeebLMAO · Answer

I know this is late, but do you still need help by any chance?

kittybasil · Answer

Actually no, it's just for reference, but if you want to do the whole thing (preferably from scratch in case I have it wrong) feel free! ~~Also I'm an idiot why did I bump this.~~

kittybasil · Answer

again, if anyone wants to contribute please do your own work cause
A) I might have done the entire thing wrong. My Physics is insanely rusty.
B) Really would prefer people don't leech/piggyback off my work! ლ(́◉◞౪◟◉‵ლ)

Timmyspu · Answer

Distance formula

$$d= (x2-x1)^2+ (y2-y1)^2$$

(x1,y1) (x2,y2) 
(9,3) (8,2)

$$(8-9)^2+(2-30)^2$$

$$-1^2+-28^2$$

1+784

square root 785

28.01

kittybasil · Answer

Bless you ლ(́◉◞౪◟◉‵ლ) (Also yes I gave the medal, in case the context of this comment is confusing)

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