### Video Transcript

A particle moves along the π₯-axis with velocity given by π£ of π‘ equals five sin of 0.8π‘ squared over two π‘ to the fourth power minus three π‘ squared plus two for time π‘ is greater than or equal zero and less than or equal to 2.5. The particle is at position π₯ equals negative four at time π‘ equals zero. One, find the acceleration of the particle at time π‘ equals two. Two, find the position of the particle at time π‘ equals two. Three, evaluate the integral between zero and 2.5 of π£ of π‘ with respect to π‘ and evaluate the integral between zero and 2.5 of the magnitude of π£ of π‘ with respect to π‘. Interpret the meaning of each integral in the context of the problem. Four, a second particle moves along the π₯-axis with position given by π₯ two equals two π‘ to the fourth power minus π‘ plus six for π‘ is greater than or equal to zero and less than or equal to 2.5. At what time π‘ are the two particles moving with the same velocity?

Our particle is moving along the π₯-axis. Itβs moving in a single straight line. And we can therefore apply what we know about calculus with relation to rectilinear motion. We have been given a function for velocity in time. And weβre looking to find the acceleration of the particle at time π‘ equals two. Remember, acceleration is given by change in velocity with respect to time. In other words, acceleration is the derivative of our function velocity with respect to time. So weβre going to need to differentiate our function π£ with respect to π‘ and then evaluate it to time π‘ equals two. And itβs highly likely that the calculator youβre using will have that functionality. But letβs remind ourselves how to differentiate this function π£ with respect to time.

The function π£ is the quotient of two separate functions. So we use the quotient rule. And this says, for two differentiable functions π’ and π£, not to be confused with π£ in our question, the derivative of π’ divided by π£ with respect to π₯ is equal to π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all over π£ squared. And since we are indeed working with velocity, weβll change our letters to π and π. The numerator of our fraction is five sin of 0.8π‘ squared. So weβll that be equal to π. And the denominator, weβll let that be equal to π. We need to differentiate each of these functions with respect to π‘.

Now, itβs fairly easy to find dπ by dπ‘. Itβs equal to four times two π‘ to the third power. Thatβs eight π‘ cubed. And then, weβre going to subtract two times three π‘ which is six π‘. So dπ by dπ‘ is eight π‘ cubed minus six π‘. Weβll need to use the chain rule to find the derivative of π with respect to π‘. We differentiate five sin of π’. And we get five cos of π’, where π’ is 0.8π‘ squared. And we multiply that by the derivative of π’, which is 1.6π‘. We can now substitute everything we know about our function into the formula for the quotient rule. And we see dπ£ by dπ‘ is as shown.

Remember, weβre looking to evaluate this when π‘ is equal to two. So we replace every instance of π‘ with two. And when we do, we see that dπ£ by dπ‘, evaluated at π‘ equals two, is negative 0.69467, and so on. And since we have no units for time nor displacement, we can say that the acceleration, correct to three significant figures, is negative 0.695.

For part two, weβre being asked to find the position of the particle at time π‘ equals two. Remember, we do need to be a little bit careful when dealing with position versus displacement. We know that we can find a function for displacement by integrating a function for velocity. And we say that displacement is equal to change in position. So weβre going to initially find the displacement of the particle and then take into account the fact that it was at π₯ equals negative four at time π‘ equals zero. The displacement of the particle in the first two seconds is given by the integral evaluated between zero and two or five sin of 0.8π‘ squared over two π‘ to the fourth power minus three π‘ squared plus two.

Typing that into our calculator, and we get 2.524, and so on. The position is therefore negative four plus this value; thatβs negative 1.4758. Correct to three decimal places, we can say that the position of the particle at time π‘ equals two is π₯ equals negative 1.476.

For part three, weβre going to use our calculator to evaluate the integral between zero and 2.5 of π£ of π‘ with respect to π‘ and the integral between zero and 2.5 of the absolute value of π£ of π‘ with respect to π‘. We should remember to replace π‘ with π₯ when we type in this into our calculator, as itβs going to integrate it with respect to π₯. And when we evaluate the integral of π£ of π‘ with respect to π‘ between zero and 2.5, we get 2.482.

Remember, thatβs the displacement of our particle in the closed interval zero to 2.5. When we evaluate the integral of the absolute value of our function, weβre finding a scalar quantity. Weβre finding the distance the particle travels over the closed interval π‘ is greater than or equal to zero and less than or equal to 2.5. And this time, we get 2.567.

For part four, a second particle moves along the π₯-axis with position given by two π‘ to the fourth power minus π‘ plus six. Weβre looking to find the time at which the particles are moving with the same velocity. Letβs clear some space to answer this.

We know that we can find the velocity of the second particle by finding the derivative for- of its function for position with respect to time. Thatβs the derivative of two π‘ to the fourth power minus π‘ plus six, which is eight π‘ cubed minus one. Since weβre looking to find the time at which the two particles are moving with the same velocity, weβre going to set this equal to the function we had for velocity for the first particle. Thatβs five sin of 0.8π‘ squared over two π‘ to the fourth power minus three π‘ squared plus two. Weβll multiply both sides by the numerator of our fraction and then use our calculator to solve this equation. And when we do, we see that, correct to three significant figures, π‘ is equal to 0.735.