Video: Finding the Initial Velocity of a Particle given the Acceleration Expression

A particle moves along the π‘₯-axis. At time 𝑑 seconds its acceleration is given by π‘Ž = (4𝑑 + 6) m/sΒ², 𝑑 β‰₯ 0. Given that at 𝑑 = 2 s, its velocity is 28 m/s, what is its initial velocity?

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Video Transcript

A particle moves along the π‘₯-axis. At a time of 𝑑 seconds, its acceleration is given by π‘Ž is equal to 4𝑑 plus six meters per second squared, when 𝑑 is greater than or equal to zero. Given that at the time 𝑑 is equal to two seconds its velocity is 28 meters per second, what is the particle’s initial velocity?

The first thing that we should notice is that our particle is moving in a straight line along the π‘₯-axis. And we’re given an equation for the acceleration of this particle. We know the acceleration is the rate of change in the velocity. What this tells us is that the acceleration of our particle is equal to the derivative of the velocity of the particle with respect to time. We could then integrate both sides of this equation with respect to 𝑑. We see that the integral of our derivative function with respect to 𝑑 would just be equal to the velocity function up to a constant of integration 𝑐.

We also know from the question that our function π‘Ž is equal to 40 plus six. So the integral of π‘Ž with respect to 𝑑 is equal to the integral of 40 plus six with respect to 𝑑. To integrate 40, we add one to the exponent and then divide by this new exponent, giving us 40 squared over two, which we can simplify to two 𝑑 squared. We also know that the integral of six with respect to 𝑑 is just 60. And then, we add our constant of integration π‘˜. Equating these two equations gives us that 𝑣 plus 𝑐 is equal to two 𝑑 squared plus 6𝑑 plus π‘˜. We can then rearrange just to see that 𝑣 must be equal to two 𝑑 squared plus 6𝑑 plus π‘˜ minus 𝑐. Since π‘˜ and 𝑐 are both just constants of integration, we can combine these into a new constant, which we will call 𝑐 one. Giving us an equation for the velocity of our particle. The velocity of our particle is equal to two 𝑑 squared plus 6𝑑 plus 𝑐 one.

If we go back to the question, we’re told that when 𝑑 is equal to two, the velocity of the particle is 28 meters per second. This tells us that 𝑣 of two must be equal to 28. So by substituting 𝑑 is equal to two into our velocity function, we see that 28 must be equal to two multiplied by two squared plus six multiplied by two plus our constant 𝑐 one. We can then rearrange and solve this to see that 𝑐 one must be equal to eight. So we have the velocity of the particle at a time 𝑑 must be equal to two 𝑑 squared plus 6𝑑 plus eight. The question asks us to find the initial velocity of the particle, which is the velocity of the particle when the time 𝑑 is equal to zero. Therefore, we can substitute 𝑑 equals zero into our velocity function to see that the initial velocity of our particle must be equal to two multiplied by zero squared plus six multiplied by zero plus eight. Which we can calculate to give us eight, which means that we have shown that the particle has an initial velocity of eight meters per second.

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