The equation of motion of the particle is given by y = 40t - 16t^2 where y is in meters and t is in seconds.

The velocity of the particle is the rate of change of y with time. This is the first derivative of y with respect to t.

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The equation of motion of the particle is given by y = 40t - 16t^2 where y is in meters and t is in seconds.

The velocity of the particle is the rate of change of y with time. This is the first derivative of y with respect to t.

`dy/dt` = 40 - 32t

At t = 2, the velocity is -24 m/s

The acceleration of the particle is the second derivative of y.

`(d^2y)/(dt^2)` = -32

**The velocity of the particle at t = 2 is -24 m/s and the acceleration of the particle is a constant -32 m/s^2.**

Hi, lovecars,

For this problem, it would help if we can take derivatives. I believe y would be considered the distance travelled in a certain amount of time. To find the velocity, take the derivative of that equation:

**y'** = (40t-16t^2)' **= 40 - 32t**

Then, for acceleration, we take the derivative of the velocity equation:

**y"** = (40 - 32t)' **= -32**

So, these would be the equations for the velocity and acceleration as a function of t:

**velocity = 40 - 32t, ***this would be in meters/second*

**acceleration = -32, ***this would be in meters/second^2*

For the acceleration equation, that simply means the acceleration is constant in this equation (no t in the equation, it dropped out).

Now, to find these values at t=2, first let's do the acceleration. We said the acceleration was constant in this problem. So, at t = 2:

**acceleration = -32 meters/second^2**

This means the particle is slowing down, possibly.

For the velocity, though, we would need to plug in 2 for t:

**velocity = 40 - 32t**

**velocity = 40 - 32*2 = 40 - 64 = -24 meters/second**

Even though we have a negative number, this means the particle is going 24 meters/second "in the opposite direction", that's all.

I hope this helps. Good Luck.