How to find the minimum value of velocity?

Question

A function of trajectory is:

y = 0.2t^(3) - 0.5t^(2) + 8

The time t is in seconds.

When in seconds does the velocity reach its minimum velocity?


y' = 0.6t^(2) - t

As some kind person has already worked out the minimum velocity occurs at t = 1.6667 seconds

However, when i plot y' = 0.6X^(2) - X on my calculator the minimum value is equal to 0.83333, which is the answer that the lecturer has achieved.

I am unsure on how to calculate the correct minimum velocity.

Can someone please show working and help me solve this.

Please show working

Thanks heaps

Answer 1

well, from a physics perspective 0.83333 s is correct



the equation gives y, the vertical height

solving the derivative at zero give (0 or 1.6667)

these are the times when the projectile is at minimum (ground level)



minimum velocity occurs at the peak of the trajectory where V is zero and that happens at half the time for the up and down trip



what you really want is t for Y/t at minimum

Answer 2

Minimum (or maximum) velocity occurs when it's acceleration is 0.

y'' = 1.2t - 1 = 0

t = .83333



Whoever worked out the minimum velocity for you did it wrong.

Answer 3

first, you find the acceleration derivitive, which would be 1.2t-1=y''. set that to equal 0, which would be .8333333, or 5/6. you plug that back into ur velocity equation, and you recieve ur results (-5/12)

Answer 4

You got velocity:



v = y' = (0.6 x t x t) - t



At minimum velocity time point, v' = y'' = 0



v' = y'' = (1.2 x t) - 1 = 0



1.2 x t = 1



t = 1/1.2 = 0.8333s