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.