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Chain Rule, Log Differentiation & Rate of Change · Question 117 of 130

Question 117

MediumOriginal question

A particle moves along a line so that its distance from a fixed point after tt seconds is s=t3−6t2+9ts = t^{3} - 6t^{2} + 9t metres. At which times is it momentarily at rest?

Options for question 117

Show the answer and explanation
CorrectNot quiteAnswer A

Correct answer: A

Explanation

  1. At rest means velocity dsdt=3t2−12t+9=0\frac{ds}{dt} = 3t^{2} - 12t + 9 = 0.

  2. 3(t−1)(t−3)=03(t - 1)(t - 3) = 0: t=1t = 1 and t=3t = 3.

Worth noting

t=0t = 0 and t=3t = 3 are when s=0s = 0, the particle at the fixed point, not at rest.

Concept
Velocity as a rate of change
Typical time
90 s
Your time
—
The rule this testsRate of change of quantities
Practise this chapter →

Ten questions from Chain Rule, Log Differentiation & Rate of Change, at your own pace.

Practise this chapter →

Ten questions from Chain Rule, Log Differentiation & Rate of Change, at your own pace.