# Video: Finding the Time Interval of Increasing Velocity from the Expression of Velocity with Time

A particle is moving in a straight line such that its velocity 𝑣 after 𝑡 seconds is given by 𝑣 = −(𝑡² − 68𝑡 + 63) m/s, 𝑡 ≥ 0. When is the velocity of the particle increasing?

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### Video Transcript

A particle is moving in a straight line such that its velocity 𝑣 after 𝑡 seconds is given by 𝑣 equals negative 𝑡 squared minus 68𝑡 plus 63 meters per second when 𝑡 is greater than or equal to zero. When is the velocity of the particle increasing?

In this question, we’ve been given a function for velocity in terms of time. And, we’re looking to work out when the velocity of that particle is increasing. Well, the velocity of something will be increasing when its acceleration is greater than zero. And so, next, we recall that acceleration is the rate of change of velocity of the object with respect to time. In derivative form, 𝑎 is the derivative of 𝑣 with respect to 𝑡.

So, we need to work out when the first derivative of our function 𝑣 with respect to 𝑡 is greater than zero. And actually, this makes a lot of sense. We know that we have some function 𝑓. When its derivative is greater than zero, it’s increasing. And, when its first derivative is less than zero, it must be a decreasing function. So, it makes sense to begin by differentiating our function 𝑣 with respect to 𝑡. That’s negative 𝑡 squared minus 68𝑡 plus 63.

Now, this is still a little bit nasty, so we’re going to separate this somewhat. We know that we can take out any constant factors and focus on differentiating the function itself. So, let’s take out this common factor of negative one. We also know that the derivative of the sum of a number of functions is equal to the sum of the derivatives of each of those functions. So, we’re going to differentiate 𝑡 squared, negative 68𝑡, and 63, individually.

And then, we recall that for real constants 𝑎 and 𝑛, the derivative of 𝑎 times 𝑥 to the power of 𝑛 is 𝑛𝑎𝑥 to the power of 𝑛 minus one. In other words, we multiply the entire expression by the exponent and then reduce that exponent by one. This means the derivative of 𝑡 squared is two 𝑡 to the power of one or two 𝑡. The derivative of negative 68𝑡 is one times negative 68𝑡 to the power of zero. Well, 𝑡 to the power of zero is one. So, its derivative is just negative 68. And, the derivative of any constant is zero. We finally distribute our parentheses and we see that our first derivative of the function 𝑣 with respect to 𝑡 is negative two 𝑡 plus 68.

All that’s left is to find the interval of increase. We set the expression for the derivative as an inequality greater than zero and solve for 𝑡. Now, we could subtract 68 from both sides, but then we end up with a negative coefficient of 𝑡. So instead, let’s add two 𝑡 to both sides. When we do, we find that 68 is greater than two 𝑡. And finally, we divide through by two. And, we see that 34 is greater than 𝑡. In other words, 𝑡 is less than 34. The velocity of the particle is increasing when 𝑡 is less than 34.

Now, let’s just go back to our inequality and double check that we would have got the same answer had we indeed subtracted 68 to start. We would have obtained negative two 𝑡 is greater than negative 68. Then, we divide both sides by negative two. But of course, when we divide or multiply through in an inequality by a negative, we must reverse the inequality symbol. And so, once again, we do indeed end up with 𝑡 being less than 34.