The maximum value of the function occurs when the slope = 0The slope is the 1st derivative of the position vs time equation.1st derivative = 112 – 2 * 16 * t = 112 – 32 * t112 – 32 * t = 0Time = 112 ÷ 32 = 3.5 secondsSubstitute this value of time into the position vs time function.Position (t) = +9 + 112 * 3.5 – 16 * 3.5^2Notice that the equation for the 1st derivative of position vs time equation is the velocity vs time equation.......
Position (t) = +9 + 112 * t + ½ * -32 * t^2
Position (t) = +9 + 112 * t – 16 * t^2
But now, there is a question that asks, "Find the instantaneous velocity at 1.5 second." Substitute 1.5 for t in the equation below.
Velocity (t) = 112 + -32 * t = 112 – 32 * t
At what time is the toy rocket at its maximum height?"
The graph of position vs time is an inverted parabola.
The maximum value of the function occurs when the slope = 0
The slope is the 1st derivative of the position vs time equation.
1st derivative = 112 – 2 * 16 * t = 112 – 32 * t
112 – 32 * t = 0
Time = 112 ÷ 32 = 3.5 seconds
Substitute this value of time into the position vs time function.
Position (t) = +9 + 112 * 3.5 – 16 * 3.5^2
Notice that the equation for the 1st derivative of position vs time equation is the velocity vs time equation.
Since the rocket’s vertical velocity decreases as the rocket rises, the velocity at the highest position must equal 0 ft/s
Velocity (t) = 112 – 32 * t
0 = 112 – 32 * t
t = 112 ÷ 32 = 3.5 seconds
The vertical velocity is 0 m/s, 3.5 seconds after the launch.
Velocity is the 1st derivative of position.
Distance is the 1st derivative of velocity.
SO
Distance is the 2nd derivative of position.
I hope this helps you!
we're doing the same thing in my calculus class. plug 1.5 into the velocity function to find the instantaneous velocity. i'm not sure about the other two :s
To fine instantaneous velocity you take the derivative of s(t), which in this case is v(t) and plug in 1.5 for t in the v(t) equation.
To find the max height you need to know that when max height is achieved the velocity of the rocket will be zero (because it is changing directions), therefore you can set v(t) = 0 and solve for t to find the time at which max height is achieved and then plug that t value back into s(t) to find what the max height actually is. A side note, in calculus setting the first derivative equal to zero will give you the equations critical value(s) which may be either max or mins, you have to figure out which it is.
