A painter working at 100 feet from the ground, drops his paint brush onto the street below. The function,
s(t)=-16t^2+100, gives the height(in feet) of the paint brush as it falls to the ground. The velocity at time t=2 seconds is given by
lim s(2)-s(t)/2-t
t-->2
Evaluate this limit to find the velocity of the paint brush at 2 seconds. (include units in your answer).
again, you typed it wrong
should be
lim ( s(2) - s(t))/(2-t) as t ---> 2
= lim (36 + 16t^2 - 100)/(2-t)
= lim (16(t^2 - 4)/(2-t)
= lim 16(t+2)(t-2)/(2-t)
= lim 16(-1)(t+2) as t --->2
= 16(-1)(4) = -64
Thank You!!
so did this help
To evaluate the given limit, you will need to substitute the value of t=2 into the function s(t) and simplify the expression.
Let's start by substituting t=2 into the function s(t)=-16t^2+100:
s(2) = -16(2)^2 + 100
= -16(4) + 100
= -64 + 100
= 36
So s(2) is equal to 36 feet.
Now plug the values into the expression lim s(2)-s(t)/2-t as follows:
lim s(2)-s(t)/2-t
t-->2
lim 36 - s(t)/2-t
t-->2
Next, we substitute the value t=2 into the expression:
lim 36 - s(2)/2-2
t-->2
lim 36 - 36/2-2
t-->2
lim 36 - 36/2-2 = 0/0
t-->2
At this point, we have an indeterminate form of 0/0. To evaluate the limit, we can simplify further using algebraic manipulation.
Notice that the expression 2-2 in the denominator is equal to zero, which will cause division by zero. To avoid this, we can rewrite the expression as (t-2)(-1), which will cancel out the (t-2) in the numerator.
lim 36 - 36/2-2
t-->2
lim 36 - 36/((t-2)(-1))
t-->2
lim 36(1/((t-2)(-1))) - 36/((t-2)(-1))
t-->2
Now, we can simplify further:
lim (36 - 36(t-2))/(t-2)
t-->2
Since the (t-2) terms cancel out:
lim 36 - 36 = 0
t-->2
Therefore, the evaluated limit is 0.
The velocity of the paint brush at 2 seconds is 0 feet per second.