# Question Video: Finding the Distance Covered by a Particle Moving with Uniform Deceleration in a Given Time Interval Mathematics

A particle is moving in a straight line such that its acceleration š = ā3 m/sĀ² and its initial velocity is 39 m/s. Find its displacement during the time interval from š” = 1 to š” = 9 seconds.

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### Video Transcript

A particle is moving in a straight line such that its acceleration š is equal to negative three meters per second squared and its initial velocity is 39 meters per second. Find its displacement during the time interval from š” equals one to š” equals nine seconds.

In order to solve this problem, we will use one of the equations of motion or SUVAT equations. š  is equal š¢š” plus half šš” squared, where š  is the displacement, š¢ is the initial velocity, š” is the time, and š is the acceleration of the particle. We need to calculate the displacement š  between š” equals one and š” equals nine.

In order to calculate this displacement š , we will firstly work out š  one, the displacement between š” equals naught and š” equals one. We will then calculate š  two, the displacement between š” equals nought and š” equals nine. Subtracting these two answers will give us our value of š . If we consider the time between š” equals zero and š” equals one, š¢ is equal to 39 meters per second. š is equal to negative three meters per second squared. š” is equal to one. And š  one is our unknown.

Substituting these values into the equation š  equals š¢š” plus half šš” squared gives us 39 multiplied by one plus a half multiplied by negative three multiplied by one squared. This is equal to 37.5. Therefore, the displacement from š” equals zero to š” equals one is 37.5 meters. In order to calculate the displacement š  two between the time š” equals zero and š” equals nine, our value for š¢, the initial velocity, is still 39. š is equal to negative three and š” is equal to nine.

Substituting these values into our equation gives us 39 multiplied by nine plus a half multiplied by negative three multiplied by nine squared. Therefore, š  two is equal to 229.5. The displacement of the particle from š” equals zero to š” equals nine is 229.5 meters.

In order to calculate š , the displacement doing a time interval from š” equals one to š” equals nine, we need to subtract š  one from š  two. š  is equal to 229.5 minus 37.5. This is equal to 192. Therefore, the displacement of the particle during the time interval is 192 meters.