Question

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.

Answer 1

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

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

Answer 2

0.33 m/s

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