The angle∅ swept by a particle moving round a circle at time t is given by ∅=t^3+3t-5. Find the acceleration at t=5s
To find the acceleration at t = 5s, we need to take the second derivative of the angle function ∅(t).
First, find the first derivative of ∅(t) with respect to t:
∅'(t) = d(∅)/dt = d(t^3+3t-5)/dt = 3t^2+3
Next, take the second derivative of ∅(t):
∅''(t) = d(∅'(t))/dt = d(3t^2+3)/dt = 6t
Now, substitute t = 5s into ∅''(t) to find the acceleration at t = 5s:
∅''(5) = 6(5) = 30
Therefore, the acceleration at t = 5s is 30.
To find the acceleration at a specific time, we need to take the second derivative of the angle function with respect to time (t).
Given that the angle function is ∅ = t^3 + 3t - 5, we'll first find the first derivative with respect to t:
d∅/dt = 3t^2 + 3.
Now, we can find the second derivative by differentiating the first derivative with respect to t:
d²∅/dt² = d/dt(3t^2 + 3).
Differentiating each term separately, we get:
d²∅/dt² = d/dt(3t^2) + d/dt(3).
Using the power rule for differentiation, which states that d/dt(t^n) = n*t^(n-1), we differentiate each term:
= 6t + 0.
Simplifying further, we have:
d²∅/dt² = 6t.
Now, we can find the acceleration by substituting t = 5s into the second derivative:
a = 6*t = 6*(5s) = 30 m/s².
Therefore, the acceleration at t = 5s is 30 m/s².