# Question Video: Finding the Distance Covered by a Particle in a Given Time Period Mathematics • Higher Education

A particle is moving in a straight line such that its displacement 𝑆 at time 𝑡 is given by 𝑆 = 1/3 𝑡³ − 1/4 𝑡² m. What is the total distance covered by the particle during the first 2 seconds? Write your answer approximate to two decimal places.

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### Video Transcript

A particle is moving in a straight line such that its displacement 𝑆 at time 𝑡 is given by 𝑆 is equal to one-third 𝑡 cubed minus a quarter 𝑡 squared meters. What is the total distance covered by the particle during the first two seconds? Write your answer approximate to two decimal places.

In this question, we are given an expression for the displacement of a particle 𝑆 in terms of its time 𝑡. We are asked to calculate the total distance covered during the first two seconds. As a result, we may initially be tempted to substitute 𝑡 equals two and 𝑡 equals zero into our expression and subtract the answers. However, this will not take into account the direction that the particle is moving within the two seconds. It will only calculate the displacement and not the total distance covered.

We will therefore begin by finding the times where the velocity of the particle 𝑉 is greater than and less than zero. We can find an expression for the velocity by differentiating our expression for the displacement with respect to 𝑡. 𝑉 is equal to d𝑠 by d𝑡. Using the power rule of differentiation, d𝑠 by d𝑡 is equal to three over three 𝑡 squared minus two over four 𝑡. And this simplifies to 𝑡 squared minus one-half 𝑡.

Setting our expression equal to zero, we can solve the equation by first factoring the quadratic. We have 𝑡 multiplied by 𝑡 minus one-half is equal to zero. This gives us two values of 𝑡, where the velocity is equal to zero, 𝑡 is equal to zero, and 𝑡 is equal to one-half. Substituting any value of 𝑡 between zero and one-half into our expression for 𝑉 gives us a negative answer. And we can therefore conclude that the velocity is less than zero when 𝑡 is greater than zero and less than one-half. In a similar way, when 𝑡 is greater than one-half and less than two, the velocity 𝑉 is greater than zero. We therefore need to find the value of 𝑆 between 𝑡 equals zero and 𝑡 equals one-half and separately between 𝑡 equals one-half and 𝑡 equals two.

Subtracting 𝑆 of zero from 𝑆 of one-half will give us the displacement in the first half a second. And taking the magnitude or absolute value of this will give us the distance covered in this time. Clearing some space, when 𝑡 is equal to one-half, 𝑆 is equal to one-third multiplied by a half cubed minus one-quarter multiplied by a half squared. This is equal to negative one over 48. When 𝑡 is equal to zero, 𝑆 is equal to zero. This means that the distance covered between 𝑡 equals zero and 𝑡 equals one-half is the absolute value of negative one forty-eighth minus zero, which in turn is equal to one forty-eighths.

We repeat this process for the time period between 𝑡 equals one-half and 𝑡 is equal to two. When 𝑡 is equal to two, 𝑆 is equal to one-third multiplied by two cubed minus one-quarter multiplied by two squared. This simplifies to eight-thirds minus one, which is equal to five-thirds. The distance traveled between 𝑡 equals one-half and 𝑡 equals two is therefore equal to the absolute value of five-thirds minus negative one forty-eighth. And this is equal to twenty-seven sixteenths.

The total distance covered by the particle during the first two seconds is therefore equal to one forty-eighth plus twenty-seven sixteenths. And this is equal to forty-one twenty-fourths. As we are asked to approximate the answer to two decimal places, the total distance covered during the first two seconds is 1.71 meters.