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Question Video: Finding the Distance Covered by a Decelerating Body Mathematics

A particle was decelerating in a straight line at a rate of 4 cm/sΒ². If it momentarily came to rest 10 seconds after it started moving, find the distance it covered in 18 seconds.

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Video Transcript

A particle was decelerating in a straight line at a rate of four centimeters per second squared. If it momentarily came to rest 10 seconds after it started moving, find the distance it covered in 18 seconds.

In order to answer this question, we will use the equations of constant acceleration, known as the SUVAT equations. 𝑠 is the displacement of the particle, 𝑒 its initial velocity, 𝑣 its final velocity, π‘Ž is acceleration, and 𝑑 is time. There are two parts to this question. And in both parts, the deceleration is four centimeters per second squared. This means that our acceleration, π‘Ž, is negative four.

As the particle comes to rest 10 seconds after moving, 𝑑 is equal to 10 and the final velocity for the first part is zero centimeters per second. We want to calculate its displacement from the origin, 𝑠 sub one. We will use the equation 𝑠 is equal to 𝑣𝑑 minus a half π‘Žπ‘‘ squared. Substituting in our values, we get 𝑠 one is equal to zero multiplied by 10 minus a half multiplied by negative four multiplied by 10 squared. Zero multiplied by 10 is zero. Multiplying negative a half by negative four and 10 squared gives us 200. 𝑠 one is, therefore, equal to 200. The displacement of the particle from the origin after 10 seconds is 200 centimeters.

We will now consider the final eight seconds of the journey, as we want to find the distance the particle covered in 18 seconds. The acceleration is still negative four centimeters per second squared. Its initial velocity now is zero centimeters per second. We’re going to calculate the displacement 𝑠 two. This time, we will use the equation 𝑠 is equal to 𝑒𝑑 plus a half π‘Žπ‘‘ squared.

𝑠 two is equal to zero multiplied by 10 plus a half multiplied by negative four multiplied by eight squared. The right-hand side simplifies to zero minus 128. 𝑠 two is, therefore, equal to negative 128. The displacement in the second part of the journey is, therefore, negative 128 centimeters. As the displacement is negative, the particle is moving back to its start point or origin.

A common mistake here would be to add our values of 𝑠 one and 𝑠 two. However, we need to calculate the distance and not the displacement. As distance must be positive, we want the magnitude or absolute value of these two values. We need to add 200 and 128. This is equal to 328. The distance covered in 18 seconds by the particle is 328 centimeters.

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