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.