# Video: Using the Given Acceleration Expression and Initial Displacement of a Particle to Find the Displacement at a Given Time

A body moves in a straight line. At time π‘ seconds, its acceleration is given by π = (7π‘ + 19) m/sΒ², π‘ β₯ 0 . Given that the initial displacement of the body is 9 m, and when π‘ = 2 s, its velocity is 27 m/s, what is its displacement when π‘ = 3 s?

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### Video Transcript

A body moves in a straight line. At time π‘ seconds, its acceleration is given by π equals 7π‘ plus 19 meters per second squared, where π‘ is greater than or equal to zero. Given that the initial displacement of the body is nine meters and when π‘ equals two seconds, its velocity is 27 meters per second. What is its displacement when π‘ equals three seconds?

Before we start this question, itβs important to remember that to get from acceleration to velocity, we need to integrate our expression. Likewise, we need to integrate to get from the velocity to the displacement.

Weβre told in this question that π the acceleration is equal to 7π‘ plus 19. This means that the velocity will be equal to the integral of 7π‘ plus 19. Integrating 7π‘ gives us 7π‘ squared over two and integrating 19 gives us 19π‘. We must remember to add the constant at the end. π£ is equal to 7π‘ squared over two plus 19π‘ plus π.

We are also told that when π‘ is equal to two, π£, the velocity, is equal to 27. Substituting these values into the equation will enable us to work out the value of π. 27 is equal to seven multiplied by two squared divided by two plus 19 multiplied by two plus π. Two squared is equal to four, four multiplied by seven is equal to 28, and 28 divided by two is equal to 14. 19 multiplied by two is equal to 38. So we are left with 27 is equal to 14 plus 38 plus π. 14 plus 38 is equal to 52. So we have 27 is equal to 52 plus π. Subtracting 52 from both sides of this equation gives us a value of π of negative 25.

This means that the expression for our velocity at time π‘ is equal to 70 squared divided by two plus 19π‘ minus 25. We now have an equation for the acceleration and the velocity.

Our final step is to find an equation for the displacement. The displacement is equal to the integral of the velocity, in this case the integral of 7π‘ squared over two plus 19π‘ minus 25. This is equal to 7π‘ cubed over six plus 19π‘ squared over two minus 25π‘ plus π.

We were told in the question that the initial displacement was nine meters. Therefore, when π‘ equals zero, π  is equal to nine. Substituting in these values allows us to calculate π. As the first three terms are all equal to zero, we can see that π is equal to nine. This means that our equation for the displacement is equal to 7π‘ cubed over six plus 19π‘ squared over two minus 25π‘ plus nine.

The final part of our question asked us to calculate the displacement when π‘ is equal to three seconds. We, therefore, need to substitute π‘ equals three into this equation. This gives us a value of π  equal to seven multiplied by three cubed divided by six plus 19 multiplied by three squared divided by two minus 25 multiplied by three plus nine. Typing this into the calculator gives us a value of 51.

Therefore, the displacement when π‘ equals three is 51 meters.