Video: Finding the Maximum Velocity of a Particle and the Distance Travelled given an Acceleration Expression

A particle started moving in a straight line. Its acceleration at time 𝑡 seconds is given by 𝑎 = (−5𝑡² + 5) m/s², 𝑡 ≥ 0. Find the maximum velocity of the particle 𝑣_(max) and the distance 𝑥 it travelled before it attained this velocity, given that the initial velocity of the particle is 0 m/s.

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Video Transcript

A particle started moving in a straight line. Its acceleration at time 𝑡 seconds is given by 𝑎 equals negative five 𝑡 squared plus five meters per square seconds, when 𝑡 is greater than or equal to zero. Find the maximum velocity of the particle 𝑣 max and the distance 𝑥 it travelled before it attained this velocity, given that the initial velocity of the particle is zero meters per second.

To answer this question, we begin by recalling that we can find an expression for the velocity by integrating the expression for acceleration with respect to time. Similarly, we can find an expression for displacement by integrating the expression for velocity with respect to time. We also need to recall that we can find a maximum by first looking for the critical points of a function and these occur at places where the derivative of that function is equal to zero or does not exist.

Well, to find the critical points and ultimately the maximum for the velocity, we want to work out when d𝑣 by d𝑡 is equal to zero. But d𝑣 by d𝑡 is, of course, acceleration. So let’s set our expression for 𝑎 equal to zero and solve for 𝑡. That’s negative five 𝑡 squared plus five equals zero. Adding five 𝑡 squared to both sides, we get five 𝑡 squared equals five. And then dividing through by five, we obtain 𝑡 squared to be equal to one. Our final step is to take the square root of both sides. And remember, we don’t need to worry about finding the positive and negative square root of one since time must be a positive value. So 𝑡 is equal to one. And we found that there’s a critical point for our function for velocity, and it occurs when 𝑡 is equal to one.

Since this is the only critical point, it’s safe to assume it probably is a maximum. But we’ll check by finding the second derivative. If the second derivative is less than zero when 𝑡 is equal to one, then we will indeed have a maximum. And of course, the second derivative of velocity with respect to time is equal to the first derivative of acceleration with respect to time. Well, the first derivative of negative five 𝑡 squared plus five is simply negative 10𝑡. We’re going to evaluate this when 𝑡 is equal to one. So that’s negative 10 times one, which is, of course, negative 10. Since negative 10 is less than zero, we do indeed have a maximum when 𝑡 is equal to one.

Now, we’re actually trying to find the maximum velocity of the particle. And we know it occurs when 𝑡 is equal to one. So let’s find the function of the velocity and evaluate it when 𝑡 equals one. We said that this was the integral of the function for acceleration. So we’re integrating negative five 𝑡 squared plus five. Well that gives us negative five 𝑡 cubed over three plus five 𝑡 plus its constant of integration 𝑐. We can work out what this constant of integration is by using the fact that the initial velocity of the particle is zero meters per second. In other words, when 𝑡 is equal to zero, 𝑣 is equal to zero. Substituting these values in gives us zero equals negative five times zero cubed over three plus five times zero plus 𝑐. And that tells us that zero is equal to 𝑐. And our final expression for velocity is negative five 𝑡 cubed over three plus five 𝑡.

We want to know the maximum velocity. And we saw that this occurs when 𝑡 is equal to one. So we’re now going to substitute 𝑡 equals one into this expression. That’s negative five times one cubed over three plus five times one, which is ten thirds. And we have the first part of our solution. The max is equal to ten thirds. And it’s important to realize that there’s no other maximum at the end point where 𝑡 is equal to zero, since there are no other turning points in the interval zero to one.

We’re now going to work out the distance 𝑥 it travelled before it attained this velocity. Now, we’ll need to be a little bit careful here. Distance is the scalar version of displacement. It’s the absolute value of displacement. We know that we can integrate our function for velocity to find a function for displacement. And then, we can evaluate that between one and zero to find the total displacement. What we’ll do to check whether this is going to give us the same value as distance is check the shape of the graph. If the curve sits purely above or purely below the 𝑥-axis, then we know that the absolute value of the displacement will be equal to the distance. Well, between zero and one, the graph does indeed sit purely above the 𝑥-axis or the 𝑡-axis in this case.

So the total distance travelled will be given by the area between the curve and the 𝑡-axis bounded by the lines 𝑡 equals zero and 𝑡 equals one. That’s simply the definite integral between zero and one of the function for velocity with respect to time. Integrating gives us negative five 𝑡 to the fourth power over 12 plus five 𝑡 squared over two. And when we evaluate that between one and zero, we get negative. five over 12 plus five over two minus zero, which is 25 12ths. 𝑣 max is equal to ten thirds and the distance 𝑥 it travelled before it attained 𝑣 max is 25 over 12 meters.

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