let s(t)= -16t^2 + 50 represent the distance in ft. from the ground of an object t seconds after initial velocity of the object? initial means t=0
1. what is the initial velocity of the object?
2. what is the velocity of the object 1 sec after it was released?
3. when does the object hit the ground?
4. what is the object's velocity as it his the ground?
1. Well, if we're talking about the initial velocity, that's the velocity when time (t) is equal to zero. So, in this case, the initial velocity of the object is... well, zero! It had to start somewhere, right?
2. Ah, one second after it was released? That's a little more interesting. To find the velocity at a specific time, we can take the derivative of the function with respect to time (t), which gives us the velocity function. So, let's find the derivative of s(t):
s'(t) = -32t
Now, we can substitute t = 1 into the velocity function:
s'(1) = -32(1) = -32 ft/s
Therefore, the velocity of the object one second after it was released is -32 ft/s! That object sure is zooming downwards.
3. To find when the object hits the ground, we need to set the distance s(t) equal to zero and solve for t:
-16t^2 + 50 = 0
If we rearrange the equation, we get:
16t^2 = 50
Now, divide both sides by 16:
t^2 = 50/16
Taking the square root of both sides:
t = ±√(50/16)
So, the object hits the ground at approximately t = ±1.58 seconds (rounded to two decimal places). We usually consider only positive time, so let's say it hits the ground at t = 1.58 seconds.
4. Ah, the velocity as the object hits the ground. Well, at the moment of impact, the velocity becomes zero. You know, because the object wants to take a moment to say "goodbye" to the ground before it comes to a stop. So, when the object hits the ground, its velocity is 0 ft/s. It's time to bring out the parachute!
To answer the questions, we need to differentiate the function s(t) with respect to t because velocity is the derivative of the position function.
1. To find the initial velocity, we need to find the derivative of s(t) when t = 0.
Differentiating s(t) = -16t^2 + 50 with respect to t:
s'(t) = -32t
When t = 0, s'(t) = -32(0) = 0
Therefore, the initial velocity of the object is 0 ft/s.
2. To find the velocity of the object 1 second after it was released, we need to find the derivative of s(t) and evaluate it at t = 1.
Differentiating s(t) = -16t^2 + 50 with respect to t:
s'(t) = -32t
When t = 1, s'(t) = -32(1) = -32
Therefore, the velocity of the object 1 second after it was released is -32 ft/s.
3. To find when the object hits the ground, we need to set s(t) equal to 0 and solve for t.
-16t^2 + 50 = 0
Adding 16t^2 to both sides:
16t^2 = 50
Dividing both sides by 16:
t^2 = 50/16
Taking the square root of both sides:
t = ± √(50/16)
Since time cannot be negative, t = √(50/16)
Therefore, the object hits the ground approximately 1.60 seconds after being released.
4. To find the object's velocity as it hits the ground, we need to find the derivative of s(t) and evaluate it at t = 1.60.
Differentiating s(t) = -16t^2 + 50 with respect to t:
s'(t) = -32t
When t = 1.60, s'(t) = -32(1.60) = -51.2
Therefore, the object's velocity as it hits the ground is -51.2 ft/s.
To find the answers to these questions, we will need to differentiate the given function with respect to time.
1. To find the initial velocity of the object, we need to find the derivative of s(t) with respect to t. The derivative represents the rate of change of distance with respect to time.
s'(t) = -32t
Since the initial velocity corresponds to t=0, we can substitute t=0 into the derivative equation to find the initial velocity.
s'(0) = -32(0)
s'(0) = 0
Therefore, the initial velocity of the object is 0 ft/s.
2. To find the velocity of the object 1 second after it was released, we can use the derivative equation.
s'(t) = -32t
Substituting t=1:
s'(1) = -32(1)
s'(1) = -32 ft/s
Therefore, the velocity of the object 1 second after it was released is -32 ft/s.
3. To find when the object hits the ground, we need to solve for t when s(t) = 0. This means the object has returned to the ground level.
-16t^2 + 50 = 0
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = -16, b = 0, and c = 50. Plugging in these values:
t = (± √(0 - 4(-16)(50))) / (2(-16))
t = (± √(0 + 3200)) / (-32)
t = (± √3200) / (-32)
t = (± 56.57) / (-32)
Since time cannot be negative in this context, we take the positive value:
t ≈ 1.76 seconds
Therefore, the object hits the ground approximately 1.76 seconds after it was released.
4. The velocity of the object as it hits the ground can be found using the derivative equation:
s'(t) = -32t
Substituting t = 1.76:
s'(1.76) = -32(1.76)
s'(1.76) ≈ -56.32 ft/s
Therefore, the object's velocity as it hits the ground is approximately -56.32 ft/s.