Video: Finding the Velocity of an Object given the Position as a Function of Time

A particle’s position varies according to 𝑥(𝑡) = 3.0𝑡² + 0.50𝑡³ m. What is the particle’s instantaneous velocity when 𝑡 = 2.0 s? What is the particle’s average velocity between 𝑡 = 1.0 s and 𝑡 = 3.0 s?

03:19

Video Transcript

A particle’s position varies according to 𝑥 as a function of 𝑡 equals 3.0𝑡 squared plus 0.50𝑡 cubed meters. What is the particle’s instantaneous velocity when 𝑡 equals 2.0 seconds? What is the particle’s average velocity between 𝑡 equals 1.0 seconds and 𝑡 equals 3.0 seconds?

In part one of this exercise, we want to solve for instantaneous velocity when 𝑡 equals 2.0 seconds. Then in part two, we’ll solve for the average velocity of the particle over a time interval centered on that same time value. This one begins at 1.0 seconds and ends at 3.0 seconds. In calculating instantaneous velocity, we can recall that it’s equal to the time derivative of the position as a function of time of our particle of interest. So in our case, the instantaneous velocity of the particle as a function of time equals 𝑑 𝑑𝑡, the time derivative of position as a function of time which we’re given. When we plug in our equation for position as a function of time and take the time derivative. This derivative results in the expression 6.0𝑡 plus 1.50𝑡 squared in units of meters per second.

We’ve arrived at a general expression for instantaneous velocity. But we want to solve for the velocity at a particular time value, when 𝑡 equals 2.0 seconds. To calculate that value, we plug in a value of 2.0 seconds for 𝑡 in our expression. When we calculate this value, to two significant figures, our result is 18 meters per second. That’s the instantaneous velocity of our particle when 𝑡 equals 2.0 seconds. Next, we move on to solving for the average velocity of our particle over the time interval 1.0 to 3.0 seconds. This average velocity will be equal to the position of our particle at 3.0 seconds minus its position at 1.0 seconds divided by the time interval 3.0 minus 1.0 seconds. Or, in the denominator, 2.0 seconds.

To solve for the position of our particle at various times, we can use the expression given to us in the problem statement. When we plug in a value of 3.0 seconds for 𝑡 and calculate the position, we find a result of 40.5 meters. Which we insert for our position when 𝑡 equals 3.0 seconds. We then do the same thing for 𝑡 equals 1.0 seconds, plugging that value in to our expression and finding a result of 3.50 meters. Which we then insert for our position when 𝑡 equals 1.0 seconds. We’re now ready to calculate the average velocity of our particle over the time interval of interest. When we do and round the result to two significant figures, we find it’s equal to 19 meters per second. So our average velocity and our instantaneous velocity, even though they’re centered on the same time values, are not the same.

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