Physics problem

To find position equations, you want to express the velocity as an integral. In the case of constant velocities, the equation is v * t + b, with b being the initial position. So the equations for A and B would be A=3t and B=5t.

B starts 6 seconds later, so by the time B moves, A is already at its position for t=6 seconds. Initial position of A=3 * 6seconds = 18m. Initial position of B = 0.The new equations are A=3t+18 and B=5t.

You want to know when the position of B equals that of A, so we set 3t+18=5t. Moving the equation around, we get 2t=18. So they are at same position at t=9 seconds, which is your answer.

>>375453
Thank you so much I looked for an answer for well over an hour and now that you explained it. I'm embarrassed to not have it understood earlier. Anyways thank you.

Oops, lemme correct my answer - You're already at time =6 when it starts, so add 6 to the time elapsed. 6+9=15 seconds, which is the correct answer.

>>375446

t1=6s--> s1=3m/s 6s = 18m

-->xs 5m/s = xs 3m/s + 18m

-->5x = 3x + 18 --> x=9

--> 9s after t1 = 15s at all

>>375459
haha yeah thank you again.

>>375460
Haha thank you too.