# Question Video: Finding the Acceleration of a Body given Its Velocity as a Function of Time Mathematics

A particle is moving in a straight line such that its velocity 𝑣 at time 𝑡 seconds is given by 𝑣 = (2𝑡² − 68) m/s, 𝑡 ≥ 0. Find the magnitude of the acceleration of the particle when its velocity is 94 m/s.

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### Video Transcript

A particle is moving in a straight line such that its velocity 𝑣 at time 𝑡 seconds is given by 𝑣 equals two 𝑡 squared minus 68 meters per second for 𝑡 is greater than or equal to zero. Find the magnitude of the acceleration of the particle when its velocity is 94 meters per second.

Here, we have an equation for 𝑣 in terms of time 𝑡 seconds. And we’re interested in calculating the acceleration of the particle. So let’s begin by recalling how we link velocity and acceleration. Acceleration is the rate of change of velocity. When we think about rate of change, we think about finding the derivative. So acceleration is the first derivative of velocity with respect to time. Let’s differentiate our expression for velocity term by term. We know that to differentiate a power term such as 𝑎𝑥 to the 𝑛th power for real constants 𝑎 and 𝑛, we multiply the entire term by the exponent and then reduce that exponent by one.

The same holds when we differentiate two 𝑡 squared. We multiply it by the power, that’s two, and then reduce the power by one. So we get two times two 𝑡 to the power of one or two times two 𝑡, which is simply four 𝑡. We also know that when we differentiate a constant, we get zero. So the derivative of negative 68 is zero. And therefore, acceleration is equal to four 𝑡 or four 𝑡 meters per square second. Now, the question asks us to find the acceleration when the velocity is equal to 94. But of course, our acceleration is in terms of time 𝑡. So what we’re going to do is to find the value of 𝑡 such that the velocity is equal to 94.

We replace 𝑣 in our original equation with 94, and that gives us 94 equals two 𝑡 squared minus 68. To solve for 𝑡, we add 68 to both sides, and then we get 162 equals two 𝑡 squared. Dividing through by two gives us 81 equals 𝑡 squared, and then we take the square root of both sides. Now, normally, we’ll look to find both the positive and negative square root. But we were told that 𝑡 has to be greater than or equal to zero. So in fact, we just find the positive square root, and we get 𝑡 is equal to nine. In other words, the velocity of the particle is 94 meters per second when the time is nine seconds.

The acceleration at this time then is found by substituting 𝑡 equals nine into our equation for 𝑎. We get four times nine, which is 36. And so the acceleration of the particle is 36 meters per square second. Now, we needn’t worry too much about this word magnitude here. When we consider the magnitude of the acceleration, we’re looking for its size. In other words, we’re looking for its absolute value. So had we achieved a negative result, we would’ve simply made it positive.