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

Question

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!!

Answer 1

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!

Answer 2

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

Answer 3

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.

Answer 4

Sorry.

v = ds / dt