help with an acceleration and distance problem

science_rules

1. The problem statement, all variables and given/known data
I asked a previous question that someone answered. They said to use two equations which i did, but it still didn't work. The problem is: a rocket sled goes 632 mil/hr and slows down to a stop in 1.40 seconds. find the negative acceleration and find the distance traveled during this negative acceleration. The equations suggested were: v = u + at for acceleration and v^2= u^2 + 2aS. I know the S means distance, and u stands for initial velocity. i converted the mil/hr --> met/sec


2. Relevant equations
i attempted to use the equations given: v = u + at and v^2= u^2 + 2aS to find the acceleration and distance and i also attempted to use the distance formula x = x initial + v initial X t + 1/2 at^2 to try to find the distance, which didnt work.
My book answers were -202 met/sec^2 acceleration and 119 met distance. i understand why acceleration is negative but don't know what i am doing wrong as far as solving.


3. The attempt at a solution
632 mil/hr --> 926.9 met/sec
632 mil/hr X 5280 ft/1 mil X 1 hr/3600 sec = 926.9 met/sec

using v = u + at = 926.9 met/sec + a1.40 sec. = -926.9met/sec / 1.40 =
-662.0 met/sec^2 acceleration??! is not even close to -202 met/sec.^2

using v^2 = u^2 + 2aS = (926.9 met/sec)^2 + 2(-662.0met/^2)S =
-859143.6 = 1324S= -859143.6/-1324 = 648.9 meters?? no where close to 119 meters.

using x = x initial + v initial X t + 1/2 at^2 = 0m + 926.9 met/sec. X 1.40 sec. + 1/2a(1.40)^2 = 1,297.6 + 1/2a1.96= -1,297.6 = 0.98 = -1,324 met/sec^2
A different answer from -662 met/sec^2 but still wrong.

using a = change in veloc. divided by change in time = 0 met/sec - 926.9 met/sec. divided by 1.40 sec. = -662.0 met/sec^2. same answer from using the v = u +at equation, and still wrong. there is something i'm not doing right, or something i'm missing, or both. i am lost as to what to do now. please help me understand step by step what i should be doing to solve these problems correctly.

danago

Is your conversion to m/s correct? I dont ever use the imperial measurement system so i dont really know the conversion factors, but according to an online converter i just used, you havent converter correctly. Try that again.

EDIT: I just realised, you converted to ft/sec not m/s. Convert the miles to METERS not FEET.

science_rules

:0 Oh!!! how stupid of me!!! i didnt notice what i did. i'll try and see if it works out correctly

science_rules

ok, i converted it correctly:
632 mil/hr X 1609.34 met/mil X 1 hr/3600 sec = 282.52 met/sec
then i used v = u + at
v = u + at = 282.52 + a1.40sec^2 = -282.52 / 1.40 = -202 met/sec^2 acceleration which is correct. but when i tried using v^2 = u^2 + 2aS it didn't work:
v^2 = u^2 +2aS = 282.52^2 + 2(-202)S = 79817 + 2(-202)S = -79817 = -404S=
-79817/-404 = 197 meters. this isn't right. the correct answer should be 119 meters.

science_rules

even using the distance formula: x = xinitial + vinitial t + 1/2 at^2 yields 197.5 meters which isnt the correct answer for distance. arrrrghhh!!! so i have the correct acceleration now but not the correct distance. hmmm.

LowlyPion

You can use the V² = 2*a*x
x = (282.52)²/2*202 = 197.5

Are you sure you have the right initial conditions?

119 implies an acceleration in 1.4 sec of 121.42 and an original velocity of 170 m/s

science_rules

i know, but 119 meters is the answer my book gave me for the total distance. there can be typos but they aren't very often. i'll keep trying to calculate.

LowlyPion

Unless there are some other conditions not mentioned by the problem statement, trust the math.

Good luck.

science_rules

My mistake!! My answer is 198 meters! i looked at the answer in the book again and it's the same. i dont know why i kept thinking it was 119 meters! Math doesn't lie! i did do it correctly :)

science_rules

Solved :)