An object dropped from restfrom the top of a tall building falls y=16t^2 feet in the first t seconds.
Find the speed of the object at t=3 seconds and confirmalgebraically. The answer is 96 feet/sec but how do i figure it out? cans omeone show me steps? thanks
The given function is
y(t) = 16 t²
The speed is the derivative of y with respect to t, i.e.
y'(t)=dy/dt=16*(2t)=32t
So the speed at t=3 seconds,
y'(t)=y'(3)=32*3=96 ft/s.
To check algebraically, we take the distance travelled at 2.99 and 3.01 seconds, and divide the difference in distance by (3.01-2.99)=0.02 seconds.
y(2.99)=143.0416 ft
y(3.01)=144.9616 ft
approximate speed
= (y(3.01)-y(2.99))/(3.01-2.99)
= (144.9616-143.0416)/(0.02)
= 96 ft/sec. checks with previous calculations
Note: if you have not done calculus before, please let me know.
Well, it seems you're trying to calculate the speed of an object that has been dropped from a tall building. To solve this problem, let's break it down step by step.
We know that the distance fallen by the object is given by the equation y = 16t^2, where y is the distance in feet and t is the time in seconds.
To find the speed of the object at t = 3 seconds, we need to calculate its instantaneous velocity at that specific time. The velocity is simply the rate of change of the distance with respect to time.
To find the velocity, we need to take the derivative of the equation y = 16t^2 with respect to t. The derivative of 16t^2 is 32t.
Now, let's substitute t = 3 into our derivative equation:
v = 32t
v = 32(3)
v = 96 feet/sec
So, the speed of the object at t = 3 seconds is indeed 96 feet/sec.
To find the speed of the object at t=3 seconds, we need to find the derivative of the position function with respect to time.
Given that the object falls y=16t^2 feet in the first t seconds, the position function can be defined as:
y(t) = 16t^2
Taking the derivative of y(t) with respect to time (t), we have:
dy/dt = 32t
To find the speed of the object at t=3 seconds, we substitute t=3 into the derivative:
dy/dt = 32 * 3 = 96 feet/sec
Thus, the speed of the object at t=3 seconds is 96 feet/sec.
To find the speed of the object at t = 3 seconds using the equation y = 16t^2, follow these steps:
Step 1: Take the first derivative of the equation with respect to time (t) to find the velocity function.
dy/dt = 2 * 16t
Step 2: Substitute t = 3 seconds into the velocity function to find the speed at t = 3 seconds.
v = 2 * 16 * 3
Step 3: Simplify the expression.
v = 96 feet/sec
Now let's confirm this algebraically.
Step 4: Use the formula for velocity, v = dy/dt, and substitute t = 3 into the equation y = 16t^2.
v = dy/dt
v = d(16t^2)/dt
Step 5: Differentiate the equation with respect to time (t).
v = 32t
Step 6: Substitute t = 3 seconds into the equation.
v = 32 * 3
Step 7: Calculate the value.
v = 96 feet/sec
Therefore, through both the calculation and algebraic methods, the speed of the object at t = 3 seconds is confirmed to be 96 feet/sec.