The position function x(t) of a particle moving along an x axis is x = 6.00 - 8.00t2, with x in meters and t in seconds. (a) At what time and (b) where does the particle (momentarily) stop? At what (c) negative time and (d) positive time does the particle pass through the origin?

Answer 1

Take the derivative to get v = -16t then find where v = 0 which is at t=0 in this case therefore

A = 0

To find where is momentarily stops you must input 0 into the original function of x there we end up with 6 -8(0)^2 and this means the position of t = 0 is where x = 6

B = 6

To find the negative time and positive time solve for t given that x = 0, since x is equal to 6 - 8t^2 subtract 6 from both for

-6 = -8t^2

Then divide by -8 for

.75 = t^2

then take the square root of both sides for

.86602 ~ t

when time is negative just multiply both sides by -1

-.86602 ~ -t

Therefore

C ~ -.86602
D ~ .86602