A body rises vertically from the earth according to the law s = 64t – 16t2. If it has lost k times its velocity in its 48 ft rise, then k =
A)1 B) 1/2 C) 1/3 D)1/9
At height 48, we have
64t-16t^2 = 48
t=1 or 3
Since the body is rising (not falling), t=1
v(1) = 64-32*1 = 32
32 = 1/2 * 64
so, k = 1/2
v = ds/dt = 64 - 32t
at t = 0 , v = 64
when s = 48
48 = 64t - 16t^2
t^2 - 4t + 3 = 0
(t-1))(t -3) = 0
t = 1 or t = 3
at t = 1, it is still rising
v = 64 - 32= 32
So 64k=32
K= 1/2
To find the value of k, we need to determine the velocity of the body at the end of its 48 ft rise.
The velocity of an object can be obtained by taking the derivative of the displacement equation with respect to time.
First, let's differentiate the equation s = 64t - 16t^2 with respect to t:
ds/dt = 64 - 32t
Now, we can find the velocity of the body at time t by substituting t into the derivative expression:
v = 64 - 32t
To find the velocity at the end of the 48 ft rise, we let s = 48:
48 = 64t - 16t^2
Rearranging the equation, we get:
16t^2 - 64t + 48 = 0
Dividing the equation by 16 to simplify, we have:
t^2 - 4t + 3 = 0
Factoring the equation, we have:
(t - 1)(t - 3) = 0
So, t = 1 or t = 3.
To find the velocity at t = 3, substitute t = 3 into the velocity equation:
v = 64 - 32(3)
v = 64 - 96
v = -32 ft/s
Since the body is moving upward, the velocity should be positive. Therefore, we consider t = 1.
Substituting t = 1 into the velocity equation:
v = 64 - 32(1)
v = 64 - 32
v = 32 ft/s
Now, let's calculate the value of k.
We know that the body has lost k times its velocity in its 48 ft rise.
If the body has lost k times its velocity, then its final velocity after the rise is v - k × v.
Substituting the values, we have:
32 - k × 32 = 48
Rearranging the equation, we get:
32(1 - k) = 48
Dividing both sides by 32, we have:
1 - k = 48/32
1 - k = 3/2
Subtracting 1 from both sides, we have:
-k = 3/2 - 1
-k = 3/2 - 2/2
-k = -1/2
Multiplying both sides by -1, we get:
k = 1/2
So, the value of k is B) 1/2.
To find the value of k, we need to determine the velocity of the body at the end of its 48 ft rise and then calculate how many times it has lost that velocity.
First, let's differentiate the equation for displacement with respect to time to find the velocity function:
v = ds/dt = 64 - 32t
Next, we need to find the time at which the body reaches a displacement of 48 ft. We can set the displacement equation equal to 48 and solve for t:
64t - 16t^2 = 48
Rearranging the equation gives us:
16t^2 - 64t + 48 = 0
Dividing through by 16:
t^2 - 4t + 3 = 0
Factoring the quadratic equation gives us:
(t - 1)(t - 3) = 0
This gives us two possible values for t: t = 1 and t = 3. However, since we are looking for the time at which the body reaches a displacement of 48 ft, we can discard t = 3 as it corresponds to the time at which the body has already passed 48 ft.
Therefore, the body reaches a displacement of 48 ft at t = 1 second.
Substituting t = 1 into the velocity function, we get:
v = 64 - 32(1)
v = 64 - 32
v = 32 ft/s
Now, we need to calculate how many times the body has lost its velocity of 32 ft/s during its 48 ft rise. Since the velocity is decreasing at a constant rate, we can simply divide the final velocity by the initial velocity:
k = 48 ft / 32 ft/s
k = 3/2
k = 1.5
Therefore, k is equivalent to 1.5, which is not one of the given answer choices (A, B, C, or D). Assuming there might be a typo in the options, the closest option would be B) 1/2.