The displacement (cm) of a point is given by s= 〖4t〗^3-3√(1+2t) . Find the instantaneous velocity and acceleration at t = 1.5 sec.
Round to 3 significant figures
s = (4t)^3 - 3√(1+2t)
ds/dt = 3(4t)^2 * 4 - 3/√(1+2t)
= 12(4t)^2 - 3/√(1+2t)
d2s/dt2 = 12*2(4t)*4 + 3/√(1+2t)^3
= 96(4t) + 3/√(1+2t)^3
Now just plug in t = 3/2 and evaluate
s(3/2) = 6^3 - 3√4 = 216-6 = 210
s'(3/2) = 12(6^2) - 3/√4 = 432-3/2 = 430.5
s''(3/2) = 96(6) + 3/√4^3 = 576+3/8 = 576.375
To find the instantaneous velocity and acceleration at a specific time, we need to calculate the derivative of the displacement function with respect to time.
First, let's find the derivative of the displacement function to get the velocity function.
Given: s = 4t^3 - 3√(1 + 2t)
To find the derivative, we will use the power rule and the chain rule.
d/dt (s) = d/dt (4t^3) - d/dt (3√(1 + 2t))
Using the power rule, the derivative of 4t^3 is:
d/dt (4t^3) = 12t^2
For the second term, to find the derivative of 3√(1 + 2t), let's simplify it first.
Let u = 1 + 2t.
Then, √(1 + 2t) = (1 + 2t)^(1/2)
Now, let's use the chain rule to find the derivative.
d/dt (3√(1 + 2t)) = d/dt (3(u)^(1/2))
= 3 * (1/2)(u)^(-1/2) * (du/dt)
Substituting u = 1 + 2t and du/dt = 2:
d/dt (3√(1 + 2t)) = 3 * (1/2)(u)^(-1/2) * 2
= 3(u)^(-1/2)
Now, we can substitute our values back into the derivative equation:
d/dt (s) = 12t^2 - 3(u)^(-1/2)
Simplifying further, since u = 1 + 2t:
d/dt (s) = 12t^2 - 3(1 + 2t)^(-1/2)
Now, we have the velocity function.
To find the instantaneous velocity at t = 1.5 sec, substitute t = 1.5 into the velocity function:
v = 12(1.5)^2 - 3(1 + 2(1.5))^(-1/2)
v ≈ 40.50 - 1.29
v ≈ 39.21 cm/s
The instantaneous velocity at t = 1.5 sec is approximately 39.21 cm/s.
To find the acceleration function, we need to take the derivative of the velocity function with respect to time.
d/dt (v) = d/dt (12t^2) - d/dt (3(1 + 2t)^(-1/2))
Using the power rule and the chain rule, we can find the derivative of each term.
d/dt (12t^2) = 24t
d/dt (3(1 + 2t)^(-1/2)) = 3(-1/2)(1 + 2t)^(-3/2) * 2
Simplifying further:
d/dt (v) = 24t - 3(1 + 2t)^(-3/2)
Now, we have the acceleration function.
To find the instantaneous acceleration at t = 1.5 sec, substitute t = 1.5 into the acceleration function:
a = 24(1.5) - 3(1 + 2(1.5))^(-3/2)
a ≈ 36 - 1.5
a ≈ 34.50 cm/s^2
The instantaneous acceleration at t = 1.5 sec is approximately 34.50 cm/s^2.
So, the instantaneous velocity at t = 1.5 sec is approximately 39.21 cm/s, and the instantaneous acceleration at t = 1.5 sec is approximately 34.50 cm/s^2.