The position equation for a particle is s(t)=√t^3+1 where "s" is measured in feet and "t" is measured in seconds. Find the acceleration of the particle at 2 seconds
A) 1ft/sec^2
B) 2/3 ft/sec^2
C) - 1/108 ft/sec^2
D) None of these
For s" I get
(3t(t^3+4)) / (4(t^3+1)^(3/2))
so, s"(2) = 2/3
if you mean
s = (t^3 + 1)^0.5
then
s' = v = 0.5 (t^3+1)^-0.5 ( 3 t^2) = 1.5 t^2 (t^3+1)^-0.5
s" = a = 1.5t^2(-0.5)(t^3+1)^-1.5 (3t^2) + 3t(t^3+1)^-0.5
at t = 2
a = -3 (9)^-(3/2) + 6(9)^-(1/2)
= -3 /3^3 + 6/3
= -1/9 + 2 = 17/9
check my arithmetic !!!!
To find the acceleration of the particle at 2 seconds, we need to take the second derivative of the position equation.
First, let's find the first derivative:
s'(t) = d(s(t))/dt = d/dt (√t^3 + 1)
To differentiate the expression, we use the chain rule:
s'(t) = 1/2 * (t^3 + 1)^(-1/2) * (3t^2)
Simplifying this further, we get:
s'(t) = 3t^2 / (2√(t^3 + 1))
Now let's find the second derivative:
s''(t) = d(s'(t))/dt = d/dt (3t^2 / (2√(t^3 + 1)))
Again, applying the chain rule:
s''(t) = (6t / (2√(t^3 + 1))) - (3t^2 / (4(t^3 + 1)^(3/2)))
Now, let's substitute t = 2 into the second derivative equation:
s''(t) = (6(2) / (2√(2^3 + 1))) - (3(2^2) / (4((2^3 + 1)^(3/2))))
Simplifying this expression will give us the acceleration at 2 seconds.
s''(2) = (12 / (2√(8 + 1))) - (12 / (4((8 + 1)^(3/2))))
s''(2) = (12 / (2√9)) - (12 / (4(27)))
s''(2) = (12 / (2*3)) - (12 / (4*27))
s''(2) = (12 / 6) - (12 / 108)
s''(2) = 2 - 1/9
s''(2) = 18/9 - 1/9
s''(2) = 17/9
Therefore, the acceleration of the particle at 2 seconds is 17/9 ft/sec^2.
So, the correct option is:
D) None of these
To find the acceleration of the particle at 2 seconds, we need to take the second derivative of the position function.
First, let's find the velocity function by taking the derivative of the position function s(t):
v(t) = d/dt (s(t)) = d/dt (√t^3 + 1)
To find the acceleration function, we need to take the derivative of the velocity function:
a(t) = d/dt (v(t))
Step 1: Differentiate the position function to find the velocity function:
v(t) = d/dt (√t^3 + 1)
To differentiate √t^3, we can rewrite it as t^(3/2):
v(t) = d/dt (t^(3/2) + 1)
Using the power rule of differentiation, we have:
v(t) = (3/2)t^(1/2) + 0
Simplifying, we get:
v(t) = (3/2)t^(1/2)
Step 2: Differentiate the velocity function to find the acceleration function:
a(t) = d/dt ((3/2)t^(1/2))
Again, using the power rule of differentiation, we have:
a(t) = (1/2)(3/2)t^(-1/2)
Simplifying further, we get:
a(t) = (3/4)t^(-1/2)
Now, substitute t = 2 seconds into the acceleration function to find the acceleration at 2 seconds:
a(2) = (3/4)(2)^(-1/2)
Simplifying, we have:
a(2) = (3/4)(2)^(-1/2)
a(2) = (3/4)(1/√2)
a(2) = (3/4)(1/√2)
a(2) = 3/(4√2)
Therefore, the acceleration of the particle at 2 seconds is 3/(4√2).
None of the given options (A, B, C) match the calculated value.