The position of a particle moving along a coordinate line is s = √63 + 6t, with s in meters and t in seconds. Find the rate of change of the particle's position at t = 3 sec.

ds/dt = 0.33 m/s

Therefore, the rate of change of the particle's position at t = 3 sec is 0.33 m/s

Step-by-step explanation:

Given;

The position function of the particle.

s (t) = √ (63+6t)

The rate of change of the particle's position = ds/dt = s (t) '

Using function of function rule.

Let u = 63+6t

s = √u

ds/dt = du/dt * ds/du

du/dt = 6

ds/du = 0.5u^ (-0.5) = 0.5/u^ (0.5) = 0.5 / (63+6t) ^ (0.5)

ds/dt = 6 * 0.5 / (63+6t) ^ (0.5)

ds/dt = 3 / (63+6t) ^ (0.5)

At t = 3sec

ds/dt = 3 / (63+6 (3)) ^ (0.5) = 3/9

ds/dt = 0.33 m/s

Therefore, the rate of change of the particle's position at t = 3 sec is 0.33 m/s

0.33 m/s

Step-by-step explanation:

Given,

s = √ (63+6t) ... Equation 1

s' = ds/dt

Where s' = rate of change of the particles position.

Differentiating equation 1,

s = (63+6t) ¹/²

s' = 6*1/2 (63+6t) ⁻¹/²

s' = 3 (63+6t) ⁻¹/²

s' = 3/√ (63+6t) ... Equation 2

At t = 3 s,

Substitute the value of t into equation 2

s' = 3/√ (63+6*3)

s' = 3/√ (63+18)

s' = 3/√ (81)

s' = 3/9

s' = 0.33 m/s.

Hence the rate of change of the particles position = 0.33 m/s