Answer:
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
Answer:
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
