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

A particle is moving in a straight line such that its acceleration π‘Ž = βˆ’3 m/sΒ² and its initial velocity is 39 m/s. Find its displacement during the time interval from 𝑑 = 1 to 𝑑 = 9 seconds.

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Video Transcript

A particle is moving in a straight line such that its acceleration π‘Ž 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 𝑑 equals one to 𝑑 equals nine seconds.

In order to solve this problem, we will use one of the equations of motion or SUVAT equations. 𝑠 is equal 𝑒𝑑 plus half π‘Žπ‘‘ squared, where 𝑠 is the displacement, 𝑒 is the initial velocity, 𝑑 is the time, and π‘Ž is the acceleration of the particle. We need to calculate the displacement 𝑠 between 𝑑 equals one and 𝑑 equals nine.

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

Substituting these values into the equation 𝑠 equals 𝑒𝑑 plus half π‘Žπ‘‘ 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 𝑑 equals zero to 𝑑 equals one is 37.5 meters. In order to calculate the displacement 𝑠 two between the time 𝑑 equals zero and 𝑑 equals nine, our value for 𝑒, the initial velocity, is still 39. π‘Ž is equal to negative three and 𝑑 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, 𝑠 two is equal to 229.5. The displacement of the particle from 𝑑 equals zero to 𝑑 equals nine is 229.5 meters.

In order to calculate 𝑠, the displacement doing a time interval from 𝑑 equals one to 𝑑 equals nine, we need to subtract 𝑠 one from 𝑠 two. 𝑠 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.

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