Question Video: Determining the Acceleration and the Velocity given the Displacement Expression of a Particle Mathematics

A particle moves along the π‘₯-axis such that, at time 𝑑 seconds, its displacement from the origin is given by π‘₯ = [15 sin (2𝑑) + 10 cos (2𝑑) + 118] m, 𝑑 β‰₯ 0. Find the particle’s velocity, 𝑣, and acceleration, π‘Ž, at 𝑑 = πœ‹ s.

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Video Transcript

A particle moves along the π‘₯-axis such that, at time 𝑑 seconds, its displacement from the origin is given by π‘₯ equals 15 sin two 𝑑 plus 10 cos two 𝑑 plus 118 meters, for 𝑑 is greater than or equal to zero. Find the particle’s velocity, 𝑣, and acceleration, π‘Ž, at 𝑑 equals πœ‹ seconds.

In this question, we’ve been given an expression for the displacement of the particle at time 𝑑 seconds. So, let’s recall the link between displacement, velocity, and acceleration. Velocity is the rate of change of displacement of an object, meaning given an expression π‘₯ for displacement, if we differentiate that expression with respect to 𝑑 time, that gives us the expression for velocity. Similarly, the acceleration is the rate of change of velocity of the object. So, it’s the first derivative of velocity with respect to time. This also means that if we differentiate our expression for displacement twice to give us the second derivative with respect to time, that gives us acceleration.

So, we’re going to differentiate our expression for displacement once to find the velocity and then once again to find the acceleration. We’ll remember, of course, that we can do this term by term. And we’ll differentiate 15 sin two 𝑑 followed by 10 cos two 𝑑 followed by 118. We recall that the derivative of sin of some constant π‘Ž times π‘₯ with respect to π‘₯ is π‘Ž cos π‘Žπ‘₯. This means then that the derivative of sin of two 𝑑 is two times cos of two 𝑑. And in turn, the derivative of 15 sin two 𝑑 is two times 15 cos two 𝑑. Similarly, differentiating cos of π‘Žπ‘₯ with respect to π‘₯ gives us negative π‘Ž sin of π‘Žπ‘₯, so the derivative of 10 cos of two 𝑑 is negative two times 10 sin of two 𝑑.

We also know that to differentiate a constant, we simply get zero. So the derivative of 118 is zero. Our expression for 𝑣 then must be 30 cos of two 𝑑 minus 20 sin of two 𝑑. And since we’re working in meters and seconds, the velocity is this expression in meters per second. Now, the question wants us to find the velocity when 𝑑 is equal to πœ‹ seconds. So, let’s substitute πœ‹ into this expression. When 𝑑 is πœ‹, velocity is 30 cos of two πœ‹ minus 20 sin of two πœ‹. Of course, sin of two πœ‹ is zero. So, 20 sin of two πœ‹ must also be zero. We also know that cos of two πœ‹ is equal to one. So, the velocity at 𝑑 equals πœ‹ seconds is 30 times one, which is simply 30. And so, velocity is 30 meters per second.

We’re now going to differentiate our expression for velocity to find an expression for acceleration. We’ll differentiate 30 cos of two 𝑑 first. And we’ll use the second derivative rule again. We get negative two times 30 sin of two 𝑑. Then, if we differentiate sin of two 𝑑, we get two cos of two 𝑑. So, negative 20 sin of two 𝑑 differentiates to negative two times 20 cos of two 𝑑. And our expression for acceleration is negative 60 sin of two 𝑑 minus 40 cos of two 𝑑. And this is in meters per square second or meters per second per second. Once again, we’re going to evaluate this when 𝑑 is equal to πœ‹. It’s negative 60 sin of two πœ‹ minus 40 cos of two πœ‹. And we know that negative 60 sin of two πœ‹ is zero and cos of two πœ‹ is one, giving us a value for π‘Ž of πœ‹ as being negative 40.

The acceleration is therefore negative 40 meters per second per second. Note also that we don’𝑑 need to be alarmed that the acceleration is a negative value. Essentially, the particle is decelerating at 𝑑 equals πœ‹ seconds; it’s slowing down.

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