Asked by kailey

Me again. One last question! Again, just needed my answer verified with any explanation or walk-through.

The position of a particle moving along a coordinate line is s=√(3+6t) with s in meters and t in seconds. find the particle's acceleration at t=1 second.

a. 1 m/sec^2

b. -1/18 m/sec^2

c. 1/3 m/sec^2

d. -1/3 m/sec^2

My answer is D.
thanks ahead of time!!
Answered by Reiny
s = (3+6t)^(1/2)

v = (1/2)(3+6t)^(-1/2) (6) = 3(3+6t)^(-1/2)

a = (-3/2)(3+6t)^(-3/2) (6)
= -9(3+6t)^(-3/2)

when t = 1
a =-9(9)^(-3/2) = -9(1/27) = -1/3

you are correct!
Answered by Damon
s = (6t+3)^.5

ds/dt = .5 (6t+3)^-.5 (6)
= 3 (6t+3)^-.5

d^2s/dt^2 = 3 [ -.5(6t+3)^-1.5 *6)
= -9 /(6t+3)^1.5
if t = 1
-9 / 9^(3/2)
-9/27
-1/3
yes D
Answered by Bosnian
a = ds / dt

a = ( 1 / 2 ) * [ 1 / sqrt ( 6 t + 3 ) ] * 6

a = 3 / sqrt ( 6 t + 3 )


t = 1

a = 3 / sqrt ( 6 * 1 + 3 )

a = 3 / sqrt ( 6 + 3 )

a = 3 / sqrt ( 9 )

a = 3 / 3

a = 1 m / s ^ 2


Answer a.







Answered by Bosnian
Sorry.

v = ds / dt
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