Video: Determining the Displacement and Velocity of a Particle given the Acceleration Expression

A particle started moving in a straight line from the origin such that its acceleration at time 𝑑 seconds is given by π‘Ž = (6𝑑 βˆ’ 2) m/sΒ², 𝑑 β‰₯ 0. Given that its initial velocity was 14 m/s, determine its velocity 𝑣 and its displacement 𝑠 when 𝑑 = 2 seconds.

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Video Transcript

A particle started moving in a straight line from the origin such that its acceleration at time 𝑑 seconds is given by π‘Ž equals six 𝑑 minus two meters per second squared, where 𝑑 is greater than or equal to zero. Given that its initial velocity was 14 meters per second, determine its velocity 𝑣 and its displacement 𝑠 when 𝑑 equals two seconds.

The velocity of the particle can be calculated by integrating the acceleration with respect to 𝑑. In a similar way, the displacement can be calculated by integrating the velocity with respect to 𝑑. In our example, to work out the velocity, we need to integrate six 𝑑 minus two. The integral of six 𝑑 is three 𝑑 squared. And the integral of two is two 𝑑. Therefore, 𝑣 is equal to three 𝑑 squared minus two 𝑑 plus 𝑐. In order to work out the value of the constant 𝑐, we need to substitute in the initial conditions. When 𝑑 equals zero, the velocity 𝑣 was equal to 14. Substituting in these values gives us a value of 𝑐 equal to 14.

This means we can calculate the velocity at any time using the equation three 𝑑 squared minus two 𝑑 plus 14. In our example, we are asked to work out the velocity when 𝑑 equals two. So we need to substitute 𝑑 equals two into the equation. This gives us 𝑣 is equal to three multiplied by two squared minus two multiplied by two plus 14. Three multiplied by two squared is equal to 12. And negative two multiplied by two is negative four. Therefore, 𝑣 is equal to 12 minus four plus 14. This gives us a value for 𝑣, when 𝑑 equals two, of 22. The velocity when 𝑑 equals two seconds is 22 meters per second.

The second part of the question asked us to work out the displacement when 𝑑 equals two seconds. In order to do this, we need to integrate three 𝑑 squared minus two 𝑑 plus 14. The integral of three 𝑑 squared is 𝑑 cubed. The integral of two 𝑑 is 𝑑 squared. And the integral of 14 is equal to 14𝑑. Therefore, 𝑠 is equal to 𝑑 cubed minus 𝑑 squared plus 14𝑑 plus 𝑐. As the particle started moving from the origin, we know that when 𝑑 is equal to zero, 𝑠, the displacement, is also equal to zero. Substituting these values into the equation gives us a value of 𝑐 equal to zero.

We can work out the displacement of the particle at any given time 𝑑 using the equation 𝑠 equals 𝑑 cubed minus 𝑑 squared plus 14𝑑. Once again, we need to substitute 𝑑 equals two into this equation. This gives us 𝑠 is equal to two cubed minus two squared plus 14 multiplied by two. Two cubed is equal to eight. And two squared is equal to four. Eight minus four plus 28 is equal to 32. This means that the displacement of the particle when 𝑑 equals two seconds is 32 meters.

After two seconds, the particle is 32 meters from the origin and is travelling with a velocity of 22 meters per second.

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