The equation of motion of a particle is given by y=40t-16t^2 where y is in meters and t is in seconds. Find the velocity as a function of t, velocity when t=2, acceleration as a function of t and acceleration at t=2.
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.
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.