Lesson Video: Linear Motion with Derivatives | Nagwa Lesson Video: Linear Motion with Derivatives | Nagwa

Lesson Video: Linear Motion with Derivatives Mathematics

In this video, we will learn how to use differentiation to find instantaneous velocity, speed, and acceleration of a particle.

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Video Transcript

In this video, we will learn how to use differentiation to find the instantaneous velocity, speed, and acceleration of a particle. We will begin by recalling some key terms, notation, and their units.

The displacement of a body, denoted 𝑠 of 𝑑, is measured in meters. The distance a body travels is the magnitude of 𝑠 of 𝑑, as the distance must always be positive. This also has a standard unit of meters. The velocity of a body is denoted 𝑣 of 𝑑 and is measured in meters per second. As with distance, the speed of a body must be positive. Therefore, the speed is equal to the magnitude of the velocity. Finally, we have the acceleration π‘Ž of 𝑑 which has a standard unit of meters per second squared.

Before we consider how we can use differentiation to solve problems, we will begin by considering a velocity–time graph and how this links these terms. The sketch shows a velocity–time graph with six distinct periods. We know that the derivative of the velocity gives us the acceleration of a body. So when the graph slopes upwards, we have a positive acceleration. And we can calculate this acceleration by working out the slope or gradient of the graph. Likewise, when the graph slopes downwards, there is a negative acceleration. Finally, when the graph is horizontal, there is no slope or gradient. This means that there is zero acceleration and the body is traveling with constant velocity.

Displacement is the change in position of an object. And we can calculate this displacement by working out the area between the graph and the 𝑑-axis. When the graph is above this axis, the displacement will be positive, and when it is below the axis, it will be negative. We will now look at a specific question involving a velocity–time graph.

The figure shows a velocity–time graph for a particle moving in a straight line. When is the acceleration of the particle negative?

We recall that when dealing with a velocity–time graph, the acceleration is negative when the slope moves downwards from left to right. This occurs during three periods on the graph. Between 𝑑 equals zero and 𝑑 equals three, the velocity 𝑣 of 𝑑 decreases from six to zero. This means it is decelerating or the acceleration is negative. From 𝑑 equals three to 𝑑 equals four, the velocity decreases from zero to negative six. And finally, between 𝑑 equals nine and 𝑑 equals 11, the velocity decreases from five to zero. We can therefore conclude that the acceleration is negative from 𝑑 equals zero to 𝑑 equals four, and it is also negative from 𝑑 equals nine to 𝑑 equals 11. There are two continuous periods on the graph where the acceleration of the particle is negative.

We will now move on to consider how we can convert between displacement, velocity, and acceleration using differentiation. We recall that the displacement of a body at any given time 𝑑 can be written as 𝑠 of 𝑑. The velocity is 𝑣 of 𝑑 and the acceleration π‘Ž of 𝑑. If we are given an expression for 𝑠 of 𝑑, we can differentiate with respect to 𝑑 to find an expression for 𝑣 of 𝑑. The same is true to move from the velocity to acceleration. This means that 𝑣 of 𝑑 is equal to d by d𝑑 of 𝑠 of 𝑑. Likewise, π‘Ž of 𝑑 is equal to d by d𝑑 of 𝑣 of 𝑑. We can also find an expression for the acceleration π‘Ž of 𝑑 by differentiating the displacement expression twice.

Whilst it will not be required in this video, we can move in the opposite direction from π‘Ž of 𝑑 to 𝑣 of 𝑑 and from 𝑣 of 𝑑 to 𝑠 of 𝑑 by integrating. This is because integration is the opposite or inverse of differentiation. We will now look at some examples where we need to find the expression for the velocity or acceleration of a body and extend this to find the velocity or acceleration at a certain time.

A particle is moving in a straight line such that its displacement 𝑠 after 𝑑 seconds is given by 𝑠 is equal to two 𝑑 squared minus three 𝑑 plus three meters, where 𝑑 is greater than zero. Determine the velocity 𝑣 as a function of time.

We are told in the question that the displacement of a particle 𝑠 is equal to two 𝑑 squared minus three 𝑑 plus three meters. And we need to find an expression for the velocity 𝑣. In order to do this, we will need to differentiate our function, as 𝑣 of 𝑑 is equal to d by d𝑑 of 𝑠 of 𝑑. If the displacement of a body is given as a function in terms of time, we can differentiate to find an expression for the velocity. Differentiating two 𝑑 squared gives us 4𝑑. Differentiating negative three 𝑑 gives us negative three. And differentiating a constant, in this case three, gives us zero. The velocity is therefore equal to 4𝑑 minus three. As we are differentiating with respect to 𝑑, 𝑣 is equal to 4𝑑 minus three meters per second. This is an expression for the velocity as a function of time.

In our next question, we need to find an expression for the acceleration as a function of time.

A particle is moving in a straight line such that its displacement 𝑠 after 𝑑 seconds is given by 𝑠 is equal to two 𝑑 cubed plus two 𝑑 plus two meters, where 𝑑 is greater than zero. Determine the acceleration π‘Ž as a function of time.

In order to answer this question, we recall that in order to find an expression for the velocity, we can differentiate an expression in terms of time for the displacement. Likewise, we can differentiate the velocity expression to calculate an expression for the acceleration. This means that 𝑣 of 𝑑 is equal to d by d𝑑 of 𝑠 of 𝑑. Likewise, π‘Ž of 𝑑 is equal to d by d𝑑 of 𝑣 of 𝑑. In this question, we are told that the displacement 𝑠 is equal to two 𝑑 cubed plus two 𝑑 plus two meters.

Differentiating this with respect to 𝑑 will give us an expression for the velocity. Differentiating two 𝑑 cubed gives us six 𝑑 squared. Differentiating two 𝑑 gives us two, and differentiating the constant two gives us zero. Therefore, 𝑣 is equal to six 𝑑 squared plus two. As the displacement is measured in meters and the time in seconds, the velocity will be measured in meters per second. 𝑣 is equal to six 𝑑 squared plus two meters per second.

Differentiating this new expression with respect to 𝑑 will give us an expression for π‘Ž. When we differentiate six 𝑑 squared, we get 12𝑑, and once again differentiating a constant gives us zero. The acceleration π‘Ž is therefore equal to 12𝑑. We can therefore conclude that if 𝑠 is equal to two 𝑑 cubed plus two 𝑑 plus two meters, then the acceleration π‘Ž of the particle as a function of time is equal to 12𝑑 meters per second squared.

In our next question, we need to calculate the magnitude of the acceleration of a particle when it is traveling at a particular velocity.

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

To answer this question, we recall that in order to find an expression for the acceleration of a particle π‘Ž of 𝑑, we differentiate the expression 𝑣 of 𝑑. The velocity of this particle at time 𝑑 seconds is equal to two 𝑑 squared minus 68 meters per second. Differentiating two 𝑑 squared gives us four 𝑑. Differentiating any constant gives us zero. Therefore, differentiating negative 68 is zero. The acceleration π‘Ž is therefore equal to four 𝑑. As the time 𝑑 is measured in seconds, the acceleration is equal to four 𝑑 meters per second squared.

We are asked to calculate the magnitude of this acceleration when the velocity is 94 meters per second. We can do this by finding the value of 𝑑 for which the velocity is 94 meters per second and then substitute this value of 𝑑 into our equation for π‘Ž of 𝑑. Setting our equation for 𝑣 equal to 94, we have two 𝑑 squared minus 68 equals 94. Adding 68 to both sides of this equation, we have two 𝑑 squared is equal to 162. We can then divide both sides of the equation by two so that 𝑑 squared is equal to 81. Square rooting both sides of this equation, we have 𝑑 is equal to the positive or negative square root of 81. As nine squared equals 81, this means that 𝑑 is equal to positive or negative nine.

We are told in the question as 𝑑 stands for time that 𝑑 is greater than or equal to zero. This means that the time at which the velocity is 94 meters per second is nine seconds. We can now substitute this value into our expression for the acceleration. π‘Ž is equal to four multiplied by nine. As four multiplied by nine is 36, we can conclude that the magnitude of the acceleration of the particle when the velocity is 94 meters per second is 36 meters per second squared.

We will now summarize the key points from this video. We saw in this video that we can use differentiation to find the velocity and acceleration of a particle as a function of time. If we have an expression for the displacement of the particle or body in terms of time, denoted 𝑠 of 𝑑, we can find an expression for 𝑣 of 𝑑, the velocity of the particle, by differentiating. Likewise, we can differentiate this expression 𝑣 of 𝑑 with respect to time to find an expression for π‘Ž of 𝑑, the acceleration. This can be written more formally as shown.

The speed of a particle is the magnitude of its velocity, and the distance is the magnitude of the displacement. This is because both speed and distance must be positive values. Whilst we didn’t cover it in this video, it is also important to note that a particle is instantaneously at rest when 𝑣 of 𝑑 is equal to zero.

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