How do I find maximum velocity of a particle?

Question

A particle moving along a straight line has position x meters at time t >0 seconds given by the formula: x=t^2-3t-1/t

Correct me if I'm wrong
v(t)= t^(-2) + 3t - 3
a(t)= -2t^(-3) + 2

My logic is that a particle will have maximum velocity when acceleration is 0 (So, the particle will not get faster). So, you set a(t)=0 and input the values of t into v(t) and the highest number is the maximum velocity. Is this logic correct?

Answer 1

You have a typo in the v(t). It should be v(t) = t^(-2) - 2t - 3
a(t) looks ok though.

To find the maximum of this, you find the critical points of v(t), which, as you said, is when a(t) is zero. You also need to examine the endpoints if there is a given interval of time. Your problem states that the formula is valid for t > 0. Since a maximum time was not stated you also need to examine the behavior as t approaches infinity and as t approaches zero from the right. It appears that in both these instances, the velocity approaches infinity. So there is no max velocity. It just keeps speeding up.

Of course, in the real world, a particle would not be able to travel faster than light. So the given formula isn't realistic.

Answer 2

You are correct. More broadly if you are trying to maximize/minize, a function f(x), you need to solve for f'(x) = 0.

Also note that once you've done what you said you need to make sure you've not found the minimum. Finally if there is a bound around the equation, e.g., find maximum such that a<=t<=b you would also need to check at t=a and b.

I just noticed that you're derivative for v(t) is wrong. I suspect i is a typo. I get:

v(t) = x'(t) = 2t -3 +t^(-2)
a(t) = v'(t) = 2 -2t^(-3)

Answer 3

v(t) = t³ - 3t² + 12t + 4 The acceleration is the 1st spinoff of the fee. call it a(t). a(t) = 3t² - 6t + 12 this could be a quadratic function, and its premier coefficient is valuable, so it might have no optimal over the area of genuine numbers. Its given area is an open era, so strictly conversing it has no endpoints and for this reason no maxima. although, this is bounded on the ends. discover the barriers of a(t) at 0 and at 3. As t ? 0+, a(t) ? 12. As t ? 3-, a(t) ? 21. The acceleration is decrease than 21 on the given area, yet we can not call 21 a optimal, by way of fact the particle's acceleration would not attain that value.