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Help with an acceleration and distance problemby science_rules
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#1
Sep1708, 09:41 PM

P: 78

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. 


#2
Sep1708, 10:07 PM

PF Gold
P: 1,131

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. 


#3
Sep1708, 10:15 PM

P: 78

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



#4
Sep1708, 10:46 PM

P: 78

Help with an acceleration and distance problem
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. 


#5
Sep1708, 10:57 PM

P: 78

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.



#6
Sep1708, 11:39 PM

HW Helper
P: 5,343

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 


#7
Sep1808, 12:13 AM

P: 78

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.



#8
Sep1808, 09:26 AM

HW Helper
P: 5,343

Good luck. 


#9
Sep1808, 10:00 AM

P: 78

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 :)



#10
Sep1808, 10:02 AM

P: 78

Solved :)



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