calculus
Me again. One last question! Again, just needed my answer verified with any explanation or walkthrough.
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!!

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!posted by Reiny

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 Dposted by Damon

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.
posted by Bosnian

Sorry.
v = ds / dtposted by Bosnian
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