A particle oscillates between the points x=40 mm and x=160 mm with an acceleration a =
k(100 - x), where a and x are expressed in mm/s2 and
respectively, and k is a constant. The velocity of the particle is 18 mm/s when x = 100 mm
and is zero at both x = 40 mm and x = 160 mm. Determine (a) the value of k,
(b) the velocity when x = 120 mm.

(a) k = 0.09 s⁻¹

(b) The velocity= ± 16.97 mm/s

Explanation:

(a) Given that the acceleration = a = k(100 - x)

Therefore;

When x = 40 mm, v = 0 mm/s hence;

At x = 100 mm, v = 18 mm/s hence we have;

k = 162/1800 = 9/100 = 0.09 s⁻¹

(b) When x = 120 mm, we have

Therefore;

v² = 2 × 144 = 288

The velocity, v = √288 = ±12·√2 = ± 16.97 mm/s.

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explanation:

answer:

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