Video: APCALC02AB-P1A-Q22-519156787864

A particle moves along the π‘₯-axis for 𝑑 β‰₯ 0 with its position at time 𝑑 given by π‘₯(𝑑) = 2𝑑³ + 3𝑑² βˆ’ 120𝑑 + 54. For which of the following values of 𝑑 is the particle at rest? [A] 𝑑 = 4 and 𝑑 = 5[B] 𝑑 = 5 [C] 𝑑 = 4 [D] 𝑑 = 0 and 𝑑 = 4

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Video Transcript

A particle moves along the π‘₯-axis for 𝑑 greater than or equal to zero with its position at time 𝑑 given by π‘₯ of 𝑑 equals two 𝑑 cubed plus three 𝑑 squared minus 120𝑑 plus 54. For which of the following values of 𝑑 is the particle at rest? a) 𝑑 equals four and 𝑑 equals five. b) 𝑑 equals five. c) 𝑑 equals four. d) 𝑑 equals zero and 𝑑 equals four.

This particle will be at rest when it is not moving, which means that the rate of change of its position with respect to time will be equal to zero. The position π‘₯ is a function of 𝑑, time. So we’re looking for the times at which dπ‘₯ by d𝑑, the rate of change of π‘₯ with respect to 𝑑, is equal to zero. To find dπ‘₯ by d𝑑, we need to differentiate the given expression with respect to 𝑑 which we can do using the power rule of differentiation.

Differentiating each term gives dπ‘₯ by d𝑑 is equal to two multiplied by three 𝑑 squared plus three multiplied by two 𝑑 minus 120. Remember the derivative of a constant is just zero. So the derivative of plus 54 is zero. This simplifies to six 𝑑 squared plus six 𝑑 minus 120. Now, we can set this derivative equal to zero and solve the resulting quadratic equation in 𝑑.

First, we can divide through by six to give the simplified quadratic 𝑑 squared plus 𝑑 minus 20 is equal to zero. We now wish to factor this quadratic. The coefficient of 𝑑 squared is one. So the first term in each bracket will just be 𝑑. And we’re looking for two numbers which sum to the coefficient of 𝑑 β€” that’s positive one β€” and multiply to the constant term of negative 20. With a bit of thought, we see that those two numbers are positive five and negative four. So the factored form of our quadratic is 𝑑 plus five multiplied by 𝑑 minus four is equal to zero.

To solve, we set each bracket in turn equal to zero. And by simple rearrangement, we find that 𝑑 is equal to negative five or 𝑑 is equal to four. If we return though to the information given in the question, we find that this equation for the position of the particle is only valid for values of 𝑑 greater than or equal to zero. So we can eliminate 𝑑 equals negative five as a solution.

We find then that the only value of 𝑑 for which the rate of change of the particle’s position with respect to time is equal to zero and hence for which the particle is at rest is 𝑑 equals four.

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