Question

The position of a particle is given by:

s(t) = t^3 + 2t^2 + 5t
t = time
s = distance
Find the acceleration at the relevant time when the velocity is 25 in/sec.

Answer 1

To find the acceleration at the relevant time when the velocity is 25 in/sec, differentiate the position function twice with respect to time. The acceleration at a velocity of 25 in/sec is 154 in/s^2.

Step-by-step explanation:

To find the acceleration at the relevant time when the velocity is 25 in/sec, we need to differentiate the given position function twice with respect to time:

s(t) = t^3 + 2t^2 + 5t

To find velocity, we differentiate once:

v(t) = 3t^2 + 4t + 5

To find acceleration, we differentiate again:

a(t) = 6t + 4

Now that we have the acceleration function, we can substitute t = 25 in/sec to find the acceleration at that velocity:

a(25) = 6(25) + 4 = 154 rac{in}{s^2}