Question
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?
Answers
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
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
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