Video: Finding the Displacement of a Particle Moving in a Straight Line given the Relation between Velocity and Time

A particle started moving, from the origin, along the π‘₯-axis. At time 𝑑 seconds, its velocity is given by 𝑣 = (1.8𝑑² + 4.7𝑑) m/s, 𝑑 β‰₯ 0 . Find its displacement, 𝑠, and acceleration, π‘Ž, at 𝑑 = 2 s.

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Video Transcript

A particle started moving, from the origin, along the π‘₯-axis. At time 𝑑 seconds, its velocity is given by 𝑣 is equal to 1.8𝑑 squared plus 4.7𝑑 meters per second, where 𝑑 is greater than or equal to zero. Find its displacement, 𝑠, and acceleration, π‘Ž, at 𝑑 is equal to two seconds.

The question tells us a particle starts moving from the origin and it moves along the π‘₯-axis. We’re also told that its velocity after 𝑑 seconds is given by 1.8𝑑 squared plus 4.7𝑑 meters per second, where 𝑑 is greater than or equal to zero. We need to use this information to find the displacement, 𝑠, and acceleration, π‘Ž, of our particle when 𝑑 is equal to two seconds. To answer this question, the first thing we need to notice is that our particle is moving along the π‘₯-axis. This means it’s moving in a straight line. We’re given the velocity of our particle at the time 𝑑. And we want to find the displacement and acceleration of our particle.

Since our particle is moving in a straight line, we can use the fact that the velocity is the rate of change of displacement with respect to time. And the acceleration is the rate of change in the velocity with respect to time. This gives us that 𝑣 is equal to d𝑠 by d𝑑 and d𝑣 by d𝑑 is equal to π‘Ž. Let’s start by finding an equation for the acceleration of our particle after 𝑑 seconds. Since π‘Ž is equal to d𝑣 by d𝑑, the acceleration is given by the derivative of 1.8𝑑 squared plus 4.7𝑑 with respect to 𝑑. And we can evaluate this derivative by using the power rule for differentiation. We multiply by the exponent and then reduce the exponent by one. This gives us 3.6𝑑 plus 4.7.

And remember, the question wants us to find the acceleration of our particle when 𝑑 is equal to two. So we substitute 𝑑 is equal to two into our equation for the acceleration. This gives us 3.6 times two plus 4.7. And we can then evaluate this to get 11.9. And since all of our units are in terms of meters and seconds and this is an acceleration, this is in terms of meters per second squared. We now need to find the displacement of our particle after two seconds. We know that 𝑣 is equal to d𝑠 by d𝑑, the derivative of 𝑠 with respect to time. And if 𝑣 is the derivative of 𝑠 with respect to 𝑑, then 𝑠 is an antiderivative of 𝑣 with respect to 𝑑.

So we can use this to find an expression for our displacement at time 𝑑 up to a constant of integration. We get 𝑠 is equal to the integral of 1.8𝑑 squared plus 4.7𝑑 with respect to 𝑑. We could then evaluate this integral by using the power rule for integration. We add one to our exponents and then divide by the new exponent. This gives us 1.8𝑑 cubed over three plus 4.7𝑑 squared over two plus the constant of integration 𝑐. We can simplify this expression. 1.8 divided by three is equal to 0.6, and 4.7 divided by two is equal to 2.35.

So we now have an expression for displacement of our particle after 𝑑 seconds. However, we don’t know the value of 𝑐. We remember the question told us that the particle started moving from the origin. In other words, when 𝑑 was equal to zero, our particle hadn’t moved. 𝑠 was equal to zero. We can use this to find the value of 𝑐. We have that 𝑠 evaluated at zero will be equal to 0.6 times zero cubed plus 2.35 times zero squared plus 𝑐. And we know that the initial displacement of our particle was zero.

We then see that 0.6 times zero cubed is equal to zero. And 2.35 times zero squared is equal to zero. So this equation simplified to give us that 𝑐 was equal to zero. So by using 𝑐 is equal to zero, we have the displacement of our particle after 𝑑 seconds, 𝑠, is equal to 0.6𝑑 cubed plus 2.35𝑑 squared.

We want to find the displacement of our particle after two seconds. So we substitute 𝑑 is equal to two into our equation for the displacement. This gives us 0.6 times two cubed plus 2.35 times two squared. And we can then calculate this expression to get 14.2. And again, since all of our units are given in terms of meters and seconds and this represents the displacement of our particle, we can write the units as meters.

Therefore, we’ve shown the displacement 𝑠 of our particle after two seconds is 14.2 meters. And the acceleration π‘Ž of our particle after two seconds is given by 11.9 meters per second squared.

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