# 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 π‘ seconds, its velocity in meters per second is given by π£ = 3 sin (10π‘) β 5 cos (8π‘), π‘ β₯ 0. What is its displacement in the interval 0 β€ π‘ β€ π/2 seconds?

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### Video Transcript

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

We begin by recalling that the displacement of a particle π  of π‘ is equal to the integral of π£ of π‘ with respect to π‘, where π£ of π‘ 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 10π‘ minus five cos eight π‘ with respect to π‘ between the limits zero and π over two.

We begin by integrating each term individually. We know that the integral of π multiplied by sin ππ‘ with respect to π‘ is equal to negative π over π multiplied by the cos of ππ‘ plus our constant of integration πΆ. Integrating three sin 10π‘ is therefore equal to negative three-tenths cos 10π‘. As we are calculating a definite integral, we donβt need to include the constant of integration.

Integrating π multiplied by cos ππ‘ with respect to π‘ gives us π over π multiplied by sin ππ‘. This means that integrating our second term gives us negative five-eighths sin of eight π‘. We need to evaluate this expression between our limits of π over two and zero.

Substituting π‘ equals π over two gives us negative three-tenths multiplied by cos of five π minus five-eighths multiplied by sin of four π. Ensuring that our calculator is in radian mode, we see that the cos of five π is equal to negative one. The sin of four π is equal to zero. This means that when π‘ is equal to π over two, our expression is equal to three-tenths. Next, we substitute π‘ 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 π‘ 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 π‘ is greater than or equal to zero and less than or equal to π over two is 0.6 meters.