Video: Studying the Movement of a Particle given Its Displacement with Respect to Time

A particle moves along the 𝑥-axis. At time 𝑡 seconds, its displacement from the origin is given by 𝑥 = (2𝑡² − 6𝑡 − 4) m, 𝑡 ≥ 0. Determine all the possible values of 𝑡, in seconds, at which the particle’s speed ‖𝑣‖ = 4 m/s.

03:09

Video Transcript

A particle moves along the 𝑥-axis. At time 𝑡 seconds, its displacement from the origin is given by 𝑥 equals two 𝑡 squared minus six 𝑡 minus four m for 𝑡 is greater than or equal to zero. Determine all the possible values of 𝑡 in seconds at which the particle’s speed magnitude of 𝑣 is equal to four meters per second.

To answer this question, we’re going to begin by recalling how we can link displacement and velocity. We know that velocity is the rate of change of displacement. So, given an expression 𝑥 for displacement, we can differentiate this with respect to time to find an expression for the velocity. Now, in fact, we’re not looking to calculate the velocity of the particle. We’re looking to find the time at which its speed is equal to four meters per second. The speed of the particle, of course, though, is the magnitude of its velocity. Another way of representing this is as the absolute value of its velocity.

So, we’ll simply begin by differentiating our expression for 𝑥 with respect to 𝑡. We can do this term by term. And we remember that to differentiate a power term, we multiply the entire term by the exponent and then reduce that exponent by one. So, the derivative of two 𝑡 squared is two times two 𝑡, which is four 𝑡. The derivative of negative six 𝑡 is negative six. And the derivative of negative four is simply zero. So, our expression for velocity is four 𝑡 minus six. We can use these vertical bars to represent the absolute value of the velocity. It’s the absolute value of four 𝑡 minus six. And the question wants us to find the values of 𝑡 such that the absolute value of four 𝑡 minus six is equal to four.

So that we can see what we need to do next, let’s plot the graph of 𝑣 equals four 𝑡 minus six. It has a slope of four, and it passes through the vertical axis at negative six. The absolute value of this looks like the pink plot. Remember, we take any negative outputs and we simply make them positive. We want to find the values of 𝑡 such that the magnitude of four 𝑡 minus six is equal to four. That’s these two locations shown. And so, there are two equations we’re going to solve. We’re going to solve the equation four 𝑡 minus six equals four. That will give us the correct value of 𝑡 for this solution.

The other solution comes from the equation negative four 𝑡 minus six equals four or negative four 𝑡 plus six equals four. Let’s solve the first equation by adding six to both sides, and we get four 𝑡 equals 10. Then, when we divide through by four, we find 𝑡 is equal to 10 over four or five over two. Similarly, we solve the second equation by first subtracting six. We get negative four 𝑡 equals negative two. Then, we divide through by negative four to get 𝑡 equals negative two divided by negative four, which is equal to one-half. And so, we can say that the values of 𝑡 at which the particle’s speed is equal to four meters per second is one-half and five over two.

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