The function gives the distances (in feet) traveled in time t (in seconds) by a particle. Find the velocity and acceleration at the given time.
s= √(t^2-5), t=3
i think i got the answer. thanks anyway..
To find the velocity and acceleration at a given time, we need to differentiate the given function with respect to time.
First, let's find the velocity by taking the derivative of the function s(t) = √(t^2 - 5) with respect to t.
Step 1: Differentiate the function with respect to t using the chain rule.
ds/dt = (1/2) * (t^2 - 5)^(-1/2) * (2t).
Step 2: Simplify the expression.
ds/dt = t / √(t^2 - 5).
Now, let's find the velocity at t = 3 by substituting t = 3 into the derivative:
v = ds/dt = 3 / √(3^2 - 5) = 3 / √4 = 3 / 2 = 1.5 feet per second.
To find the acceleration, we need to differentiate the velocity function with respect to t.
Step 1: Differentiate the velocity function with respect to t.
d^2s/dt^2 = d/dt (t / √(t^2 - 5)).
Step 2: Apply the quotient rule.
d^2s/dt^2 = ((1 * √(t^2 - 5)) - (t * (1/2) (t^2 - 5)^(-1/2) * 2t)) / (t^2 - 5).
Step 3: Simplify the expression.
d^2s/dt^2 = (√(t^2 - 5) - t^2) / (t^2 - 5).
Now, let's find the acceleration at t = 3 by substituting t = 3 into the second derivative:
a = d^2s/dt^2 = (√(3^2 - 5) - 3^2) / (3^2 - 5) = (4 - 9) / 4 = -5 / 4 = -1.25 feet per second squared.
Therefore, at t = 3, the velocity of the particle is 1.5 feet per second and the acceleration is -1.25 feet per second squared.