1. The problem statement, all variables and given/known data a car accelerates at 1.5ms^-2 from 0 ms^-1 to 30 ms^-1. It then travels at a constant speed of 30 ms^-1. 2.A bike accelerates at 2.5ms^-2 from 0ms^-1 to 20 ms^-1 and then travels at a constant speed of 20 ms^-1 find the instant where there together. (Both have the same displacement) 2. Relevant equations all constant acceleration equations 3. The attempt at a solution for part one, i can represent the distance as 300+30t. t=(30-0)/1.5=20s S=1/2(20)(30) + 30t= 300 +30t for part 2. t=(20-0)/2.5=8s S=1/2(8)(20) + 20t= 80 + 20t 80+20t=300+30t -220=10t t=-22s why is my answer coming in negative. My answer -22 is right but its negative.