Video: Determining the Time Interval during Which a Particle’s Velocity Increases given Its Displacement with Respect to Time

Alex Cutbill

A particle is moving in a straight line such that its displacement 𝑠 after 𝑡 seconds is given by 𝑠 = (3𝑡³ − 54𝑡² + 38𝑡) m, 𝑡 ≥ 0. Determine the time interval during which the velocity of the particle is increasing.

03:26

Video Transcript

A particle is moving in a straight line such that its displacement 𝑠 after 𝑡 seconds is given by 𝑠 equals three 𝑡 cubed minus 54𝑡 squared plus 38𝑡 meters, where 𝑡 is greater than or equal to zero. Determine the time interval during which the velocity of the particle is increasing.

This question asks us something about the velocity of the particle. We have the displacement of this particle 𝑠 in terms of time 𝑡. So all we have to do to find the velocity of the particle is to differentiate this with respect to time. Okay, so let’s differentiate.

Calling the velocity 𝑣, 𝑣 is the derivative with respect to time 𝑡 of displacement 𝑠. And 𝑠 is three 𝑡 cubed minus 54𝑡 squared plus 38𝑡. So differentiating term by term using the fact that the derivative 𝑑 by 𝑑𝑡 of 𝑎 times 𝑡 to the 𝑛 is 𝑎 times 𝑛 times 𝑡 to the 𝑛 minus one, we get that the velocity is 9𝑡 squared minus 108𝑡 plus 38. Now that we have the velocity, we can start to consider the problem we’ve been given to determine the time interval during which this velocity is increasing.

One way to do this would be to sketch the graph of 9𝑡 squared minus 108𝑡 plus 38 again it’s a 𝑡. Hopefully, you know how to do this. But there is another way we can solve this. The velocity of the particle is increasing if the rate of change of velocity with respect to time — that is 𝑑𝑣 by 𝑑𝑡 — is greater than zero. More generally, a general function 𝑓 is increasing when the derivative 𝑓 prime of 𝑥 is greater than zero.

So let’s find 𝑑𝑣 by 𝑑𝑡 by differentiating again. Again, we differentiate term by term using the power rule to get 18𝑡 minus 108. So when is 𝑑𝑣 by 𝑑𝑡 greater than zero? Well, 𝑑𝑣 by 𝑑𝑡 is 18𝑡 minus 108. So this is when 18𝑡 minus 108 is greater than zero, which is when 18𝑡 is greater than 108, which is when 𝑡 is greater than six.

Writing this in interval notation, this is the open interval from six to infinity, including neither endpoint and so using a parenthesis rather than square brackets. This is the time interval during which the velocity of the particle whose displacement is given by 𝑠 equals three 𝑡 cubed minus 54𝑡 squared plus 38𝑡 meters, where 𝑡 is greater than zero is increasing.

This corresponds to where the derivative of velocity with respect to time is greater than zero. And of course, we know the derivative of a velocity with respect to time by another name. It’s the acceleration of the particle. And so another way of phrasing this question would be to ask where the acceleration of the particle is greater than zero.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.