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Using this equation, when is the particle at rest? (Velocity=0)

Using this equation, when is the particle at rest? (Velocity=0)-example-1
User FredMan
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The given equation is expressed as

s(t) = t^3 - 5t^2 + 3t

This is a position function. To find the velocity function, we would differentiate the given function with respect to t. It becomes

v(t) = 3t^2 - 10t + 3

To find the time when the velocity is zero, we would substitute v(t) = 0 into the equation, It becomes

3t^2 - 10t + 3 = 0

This is a quadratic equation. We would solve for t by applying the method of factorisation. The first step is to multiply 3t^2 with 3. It becomes 9t^2. We would find two terms such that their sum or difference is - 10t and their product is 9t^2. The terms are - t and - 9t. By replacing -10t with - t - 9t, we have

3t^2 - t - 9t + 3 = 0

By factorising by grouping, we have

t(3t - 1) - 3(3t - 1) = 0

Since 3t - 1 is common, it becomes

(3t - 1)(t - 3) = 0

3t - 1 = 0 or t - 3 = 0

3t = 1 or t = 3

t = 1/3 or t = 3

The particle is at rest at t = 1/3 and t = 3

User Tofandel
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