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A particle has a position defined by the equations π‘₯ = 5𝑑² + 4𝑑 and 𝑦 = 3𝑑 βˆ’ 2. Find the speed of the particle at 𝑑 = 2.

02:18

Video Transcript

A particle has a position defined by the equations π‘₯ equals five 𝑑 squared plus four 𝑑 and 𝑦 equals three 𝑑 minus two. Find the speed of the particle at 𝑑 equals two.

Here we’ve been given a pair of parametric equations to describe the position of the particle. This means that, given a value of 𝑑, we obtain an output which is a coordinate pair for the position of the particle at that time. We could treat these as completely separate entities. Now, we’re looking to find the speed of the particle at 𝑑 equals two. Well, we know that speed is the magnitude of the velocity. And we also recall that we can find a function for velocity by differentiating a function for displacement. We might then choose to say that, in vector terms, the position of the particle at time 𝑑 is five 𝑑 squared plus four 𝑑 𝑖 plus three 𝑑 minus two 𝑗. We could then differentiate our function for position and find a function for velocity. And in fact, we can achieve this by differentiating individually the horizontal and vertical components for our displacement or our position.

We’ll begin by differentiating the horizontal component. That’s five 𝑑 squared plus four 𝑑. The derivative of five 𝑑 squared is 10𝑑. And the derivative of four 𝑑 with respected 𝑑 is simply four. Let’s repeat this process for the vertical component. The derivative of three 𝑑 is three and the derivative of negative two is zero. In vector terms then, our velocity is described by 10𝑑 plus four 𝑖 plus three 𝑗. All this means is, at 𝑑 seconds, the velocity is 10𝑑 plus four in the horizontal direction and three in the vertical, the 𝑗 direction. Next, we want to work out the velocity at 𝑑 equals two. So we’ll substitute 𝑑 into our vector equation for velocity that gives us 10 times two plus four 𝑖 plus three 𝑗. That’s 24𝑖 plus three 𝑗.

Remember that we’re looking to find the speed, which is the magnitude of the velocity. So we recall that we can find the magnitude of a vector π‘Ž given as π‘₯ 𝑖 plus 𝑦 𝑗 by finding the square root of π‘₯ squared plus 𝑦 squared. In this case, that’s the square root of 24 squared plus three squared, which is equal to three root 65. Now, we don’t have any units here, so we’re done. The speed of the particle at 𝑑 equals two is three root 65.

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