Question Video: Finding the Displacement in the given Interval of Time and Velocity Expression Mathematics • Higher Education

A particle moves in a straight line. At time 𝑑 seconds, its velocity in meters per second is given by 𝑣 = 3 sin (10𝑑) βˆ’ 5 cos (8𝑑), 𝑑 β‰₯ 0. What is its displacement in the interval 0 ≀ 𝑑 ≀ πœ‹/2 seconds?

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Video Transcript

A particle moves in a straight line. At time 𝑑 seconds, its velocity in meters per second is given by 𝑣 is equal to three sin 10𝑑 minus five cos eight 𝑑, where 𝑑 is greater than or equal to zero. What is its displacement in the interval 𝑑 is greater than or equal to zero and less than or equal to πœ‹ over two seconds?

We begin by recalling that the displacement of a particle 𝑠 of 𝑑 is equal to the integral of 𝑣 of 𝑑 with respect to 𝑑, where 𝑣 of 𝑑 is an expression for the velocity in terms of time. When calculating the displacement in a time interval, the start and end times will be our lower and upper limits, respectively. In this question, the displacement will be equal to the integral of three sin 10𝑑 minus five cos eight 𝑑 with respect to 𝑑 between the limits zero and πœ‹ over two.

We begin by integrating each term individually. We know that the integral of π‘Ž multiplied by sin 𝑏𝑑 with respect to 𝑑 is equal to negative π‘Ž over 𝑏 multiplied by the cos of 𝑏𝑑 plus our constant of integration 𝐢. Integrating three sin 10𝑑 is therefore equal to negative three-tenths cos 10𝑑. As we are calculating a definite integral, we don’t need to include the constant of integration.

Integrating π‘Ž multiplied by cos 𝑏𝑑 with respect to 𝑑 gives us π‘Ž over 𝑏 multiplied by sin 𝑏𝑑. This means that integrating our second term gives us negative five-eighths sin of eight 𝑑. We need to evaluate this expression between our limits of πœ‹ over two and zero.

Substituting 𝑑 equals πœ‹ over two gives us negative three-tenths multiplied by cos of five πœ‹ minus five-eighths multiplied by sin of four πœ‹. Ensuring that our calculator is in radian mode, we see that the cos of five πœ‹ is equal to negative one. The sin of four πœ‹ is equal to zero. This means that when 𝑑 is equal to πœ‹ over two, our expression is equal to three-tenths. Next, we substitute 𝑑 is equal to zero. This gives us negative three-tenths multiplied by cos of zero minus five-eighths multiplied by sin of zero. The cos of zero equals one and sin of zero is equal to zero. This means that when 𝑑 is equal to zero, our expression is equal to negative three-tenths.

To evaluate the displacement, we need to subtract negative three-tenths from three tenths. This is equal to six-tenths, which written as a decimal is 0.6. The displacement of the particle in the interval 𝑑 is greater than or equal to zero and less than or equal to πœ‹ over two is 0.6 meters.

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