The angle∅ swept by a particle moving round a circle at time t is given by ∅=t^2+3t-5. Find the acceleration at t=5s
To find the acceleration at t=5s, we need to find the second derivative of the angle function ∅ with respect to time t.
Taking the derivative of ∅ with respect to t, we get:
d∅/dt = 2t + 3
Taking the derivative of d∅/dt with respect to t, we get:
d²∅/dt² = 2
The acceleration at t= 5s is the value of the second derivative of ∅ at t=5s:
d²∅/dt² = 2
Therefore, the acceleration at t=5s is 2.
To find the acceleration at t = 5s, we need to calculate the second derivative of the angle function with respect to time.
Given: θ = t^2 + 3t - 5
First, let's find the first derivative:
θ' = dθ/dt = 2t + 3
Now, let's find the second derivative:
θ'' = d²θ/dt² = d/dt(2t + 3)
= 2
Since the second derivative is a constant value of 2, the acceleration at t = 5s is 2.