The position equation for the movement of a particle is give by s=(t^2)^3 when s is measured in feet and t is?

Question

measured in seconds. Find the acceleration at two seconds.
please explain steps. thank you.

Answer 1

Position:
s = (t^2)^3 = t^(2*3) = t^6 ft

Velocity:
v = ds/dt = 6*t^5 ft/s

Acceleration:
a = dv/dt = 30*t^4 ft/s/s

a(t=2) = 30*2^4 = 30*16 = 480 ft/s/s

Answer 2

Okay. Let's simplify this term first. (t^2)^3 is the same thing as saying t^6. So, s = t^6. Now, the acceleration is the second derivative of the position equation with velocity being the first derivative. So, let's compute that.

Velocity: s' = 6t^5
Acceleration: s'' = 30t^4. Now, plug in 2 for the t.
s'' = 30 * 2 ^ 4 = 30 * 16 = 480.

The acceleration at t = 2 seconds is 480 ft/sec^2

Answer 3

Start by computing the second derivative:Because the second derivative is the same as acceleration--s``=30s^4
and s``(2)=480 ft/sec^2

Answer 4

position = s = (t^2)^3
= (t^2)(t^2)(t^2)
= t^6
acceleration = s'' (2nd derivative)
s'= velocity = first derivative = 6t^5
s''=acceleration = 30t^4
@ t=2,
30(2)^4 = acceleration

Answer 5

When you substitute 2 for t, you'll get a distance. Now find an equation that expresses 'a' in terms of s and t.