The position of a particle moving along a line is given by s(t) = 2t³ - 24t² + 90t + 7 for t ≥ 0. For what values of t is the speed of the particle increasing.

Solution:

s(t) = 2t³ - 24t² + 90t + 7 and t ≥ 0

The maximum value of t for which the speed is maximum

ds(t)/dt = 6t² - 48t + 90

Equating the above to zero we have:

6t² - 48t + 90=0

t² - 8t + 15 = 0

The factors of 15 which are used for factoring are 3 and 5

t² - 3t - 5t + 15 = 0

t(t - 3) - 5(t - 3) = 0

(t - 3)(t -5) = 0

t = 3 and t =5

The second differential

d²s(t)/dt² = 2t - 8

At t = 3

d²s(t)/dt² = 2t - 8 = 6 - 8 = -2

After t=3 the speed decreases.

At t = 5

d²s(t)/dt² = 2t - 8 = 10 - 8 = 2

At t = 5 the speed further increases

From t = 0 to t = 3 the speed increases

From t = 3 to t = 5 the s= 5peed decreases

From t = 5 onward speed increases

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The position of a particle moving along a line is given by s(t) = 2t³ - 24t² + 90t + 7 for t ≥ 0. For what values of t is the speed of the particle increasing.

Summary:

The values of t for which the speed of the particle is increasing t = 1, 2 and 5 onwards