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Question Video: Determining the Displacement of a Particle at the given Velocity Using Its Acceleration Mathematics • Higher Education

A body started moving along the π‘₯-axis from the origin at an initial speed of 10 m/s. When it was 𝑠 meters away from the origin and moving at 𝑣 m/s, its acceleration was (45𝑒)^(βˆ’π‘ ) m/sΒ² in the direction of increasing π‘₯. Determine 𝑠 when 𝑣 = 11 m/s.

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Video Transcript

A body started moving along the π‘₯-axis from the origin at an initial speed of 10 meters per second. When it was 𝑠 meters away from the origin and moving at 𝑣 meters per second, its acceleration was 45 times 𝑒 to the power of negative 𝑠 meter per square second in the direction of increasing π‘₯. Determine 𝑠 when 𝑣 equals 11 meters per second.

Let’s begin by defining the acceleration to be π‘Ž such that π‘Ž is equal to 45𝑒 to the power of negative 𝑠 meters per square second. We’re given information about the speed of the body at various points during its motion. And so we begin by recalling how we can link acceleration with velocity. In particular, the acceleration is the rate of change of velocity. In other words, acceleration is the derivative of velocity with respect to time.

Now, of course, we’re interested in the initial speed, and this is the velocity. But since speed is the magnitude of velocity and the object is moving in the direction of increasing π‘₯, then these are going to be the same thing. So we can begin to answer this question by replacing π‘Ž with d𝑣 by d𝑑 in our earlier equation. So d𝑣 by d𝑑 is equal to 45 times 𝑒 to the power of negative 𝑠. And then what we might look to do is find an expression for 𝑣 here by integrating.

However, we notice that we have d𝑣 by d𝑑, and then our expression for d𝑣 by d𝑑 is actually in terms of 𝑠. And so we recall that we can alternatively represent d𝑣 by d𝑑 as d𝑣 by d𝑠 times d𝑠 by d𝑑. This, of course, is essentially a special version of the chain rule. And this allows us to rewrite our equation even further as d𝑣 by d𝑠 times d𝑠 by d𝑑 equals 45𝑒 to the power of negative 𝑠. But we also know that we can link velocity with displacement. Velocity is said to be the rate of change of displacement. In other words, velocity is the derivative of displacement with respect to time. And so our equation is now d𝑣 by d𝑠 times 𝑣 equals 45𝑒 to the power of negative 𝑠.

We now notice we have d𝑣 and 𝑣 and d𝑠 and 𝑠. And so what we’re going to do is treat d𝑣 and d𝑠 as differentials. This allows us to separate our equation and then integrate both sides. In particular, we’re allowed to treat d𝑣 by d𝑠 here a little bit like a fraction, allowing us to multiply through by d𝑠. Then, we have an equation that we can begin to solve by integrating both sides with respect to their respective variables. In other words, we’re going to integrate 𝑣 with respect to 𝑣 and 45𝑒 to the power of negative 𝑠 with respect to 𝑠.

Now, of course, the indefinite integral of 𝑣 with respect to 𝑣 is a half 𝑣 squared plus some constant of integration 𝑐. But actually we’re going to create a constant of integration on the right-hand side as well. And what we’ll actually do is just combine them into one final constant. We also know that if we integrate 𝑒 to the power of negative 𝑠, we get negative 𝑒 to the power of negative 𝑠. And so the right-hand side of our equation becomes negative 45𝑒 to the power of negative 𝑠 plus 𝑐.

And this is really useful because we now know that we have an equation for 𝑣 in terms of 𝑠. And if we rearrange this, we can determine the value of 𝑠 for some 𝑣 if we know the value of 𝑐. But of course we can calculate the value of 𝑐 by using the fact that the initial speed of the particle is 10 meters per second. In particular, when 𝑠, the displacement, is equal to zero, we can say that 𝑣 is equal to 10. Substituting this into our equation, we get a half times 10 squared equals negative 45𝑒 to the power of negative zero plus 𝑐. This simplifies to 50 equals negative 45 plus 𝑐, and we can then solve for 𝑐 by adding 45. So 𝑐 is 95. And so we have the equation that links 𝑣 with 𝑠.

We’re now going to substitute 𝑣 equals 11 into this equation and rearrange to make 𝑠 the subject. This gives us a half times 11 squared equals negative 45𝑒 to the power of negative 𝑠 plus 95. A half times 11 squared then we subtract 95 gives us negative 69 over two. This is equal to negative 45𝑒 to the power of negative 𝑠. So we’re next going to divide through by negative 45. And we get 23 over 30 equals 𝑒 to the power of negative 𝑠.

To solve this equation for 𝑠, let’s take the natural logarithm of both sides. We’ll clear some space when we do so. The left-hand side becomes the natural logarithm of 23 over 30. And then the natural logarithm of 𝑒 to the power of negative 𝑠 is simply negative 𝑠. We multiply through by negative one. So 𝑠 is negative the natural logarithm of 23 over 30. And we can simplify this a little bit further if we think about this as negative one times the natural logarithm of 23 over 30. Using the laws of logarithms, it allows us to write that left-hand side as the natural logarithm of 23 over 30 to the power of negative one. And then 23 over 30 to the power of negative one is 30 over 23. So the left-hand side simplifies even further to the natural logarithm of 30 over 23.

And so we found the displacement of the object when 𝑣 equals 11 meters per second. In meters, it’s the natural logarithm of 30 over 23.

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