Given v = 2t3 + 87t2 - 636, determine the acceleration for t= 1 s.
acceleration in m/s2 =
My Answer
v'= 6t^2+174t
V"= 12t
V"(1) = 12
acceleration in m/s2 = 12
Can you please confirm this is correct..
Thanks!
V' = 6t^2 + 174t
V" = 12t + 174.
To find the acceleration, we need to take the second derivative of the given equation for velocity, v(t).
Given v(t) = 2t^3 + 87t^2 - 636, we can find the derivative, v'(t), which represents the velocity equation:
v'(t) = d/dt [2t^3 + 87t^2 - 636]
= 6t^2 + 174t
Now, taking the second derivative, v''(t), will give us the acceleration equation:
v''(t) = d/dt [6t^2 + 174t]
= 12t + 174
To find the acceleration at t = 1s, we substitute t = 1 into the acceleration equation:
v''(1) = 12(1) + 174
= 12 + 174
= 186 m/s^2
Therefore, the acceleration at t = 1 second is 186 m/s^2. Your answer of 12 m/s^2 seems incorrect.