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!!

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!

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

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.

Sorry.

v = ds / dt

To find the particle's acceleration at t=1 second, we need to take the second derivative of the position function with respect to time.

Given the position function s = √(3 + 6t), we first find the first derivative of s with respect to t:

ds/dt = (1/2)(3 + 6t)^(-1/2) * 6

Simplifying, we get:

ds/dt = 3/(2√(3 + 6t))

Now, let's find the second derivative of s. We differentiate ds/dt with respect to t:

d^2s/dt^2 = d/dt [3/(2√(3 + 6t))]

To simplify further, let's rewrite the expression as:

d^2s/dt^2 = 3/(2(3 + 6t)^(3/2))

Finally, substitute t = 1 into the above expression to find the acceleration at t=1 second:

d^2s/dt^2 = 3/(2(3 + 6(1))^(3/2))
= 3/(2(3 + 6)^(3/2))
= 3/(2(9)^(3/2))
= 3/(2(27))
= 3/54
= 1/18

Therefore, the particle's acceleration at t=1 second is 1/18 m/sec^2.

Hence, the correct answer is option b) -1/18 m/sec^2.