Video: Verifying Whether a Given Subtraction Equation Is True or False

A moving particle is defined by the two equations π‘₯ = 𝑑³ βˆ’ 5𝑑 βˆ’ 5 and 𝑦 = 7𝑑² βˆ’ 3. Find the magnitude of the acceleration of the particle at 𝑑 = 1.

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Video Transcript

A moving particle is defined by the two equations π‘₯ equals 𝑑 cubed minus five 𝑑 minus five and 𝑦 equals seven 𝑑 squared minus three. Find the magnitude of the acceleration of the particle at 𝑑 equals one.

In this question, we’ve been given the position of the particle as defined by a pair of parametric equations. So given a value of 𝑑, we obtain a coordinate pair π‘₯𝑦 for the position of our particle. We could choose to consider this in vector terms and say that the position of the particle at time 𝑑, 𝑠 of 𝑑, is given by 𝑑 cubed minus five 𝑑 minus five 𝑖 plus seven 𝑑 squared minus three 𝑗. Now, in this question, we’re looking to find the acceleration. Well, in fact, we want the magnitude of the acceleration, but we’ll deal with that in a moment. So we recall that the acceleration is equal to the derivative of the velocity with respect to time. But we also know that the velocity is equal to the first derivative of the displacement or the position vector.

We can in turn say that, to find a function for acceleration, we’re going to need to differentiate our function for position twice with respect to time. We’ll do it once to find the function for velocity. We can differentiate each component function in turn. When we differentiate 𝑑 cubed minus five 𝑑 minus five, we get three 𝑑 squared minus five. And when we differentiate seven 𝑑 squared minus three, we get 14𝑑. So our function for velocity is three 𝑑 squared minus five 𝑖 plus 14𝑑 𝑗. But what does this actually mean? Well, it means that the velocity can be defined in terms of its horizontal velocity and its vertical velocity. Horizontally, its velocity is three 𝑑 squared minus five, but vertically, it’s given by the function 14𝑑.

Okay, great. Let’s differentiate again to find our function for acceleration. And when we differentiate three 𝑑 squared minus five, we find that the acceleration in the horizontal direction is six 𝑑. We then differentiate 14𝑑. And we see that the acceleration in the vertical direction is 14. We’re now able to find a vector acceleration for our particle at 𝑑 equals one. We simply substitute 𝑑 equals one into a vector function for acceleration. And we find that the acceleration at 𝑑 equals one is given by the vector six 𝑖 plus 14𝑗. We, of course, want to find the magnitude of the acceleration though.

So we recall that the magnitude of a vector in two dimensions given by π‘₯ 𝑖 plus 𝑦𝑗 is the square root of π‘₯ squared plus 𝑦 squared. This means the magnitude of our acceleration at 𝑑 equals one is the square root of six squared plus 14 squared, which is two root 58. There are no units here. So we found the magnitude of the acceleration of the particle at 𝑑 equals one to be two root 58.

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