Use the position function s(t)=-4.9t^2+ 150, which gives the height (in meters) of an object that has fallen from 150 meters. Teh velocity at time t=a seconds is given by: lim as t approaches a s(a)-s(t) divided by a-t
a) find the velocity of the object when t=3
b) at what velocity will the object impact the ground?
Please help.. My teacher didn't explain this very well at all.
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To solve these problems, we need to calculate the velocity using the given position function and apply the concept of limits.
a) To find the velocity when t = 3 seconds, we need to consider the limit expression:
lim as t approaches a (s(a) - s(t)) / (a - t)
In this case, a = 3 seconds. Substitute t = 3 into the position function s(t) = -4.9t^2 + 150:
s(3) = -4.9(3)^2 + 150
= -4.9(9) + 150
= -44.1 + 150
= 105.9
Now we calculate the limit expression:
lim as t approaches 3 (s(3) - s(t)) / (3 - t)
Substitute in the values:
lim as t approaches 3 (105.9 - s(t)) / (3 - t)
To simplify, we can rewrite the position function as:
s(t) = -4.9t^2 + 150
= 150 - 4.9t^2
Substituting this back into the limit expression:
lim as t approaches 3 (105.9 - (150 - 4.9t^2)) / (3 - t)
Simplify inside the brackets:
lim as t approaches 3 (105.9 - 150 + 4.9t^2) / (3 - t)
lim as t approaches 3 (-44.1 + 4.9t^2) / (3 - t)
Now we substitute t = 3 into the expression:
(-44.1 + 4.9(3)^2) / (3 - 3)
(-44.1 + 4.9(9)) / (0)
Simplify further:
(-44.1 + 44.1) / (0)
0 / 0
At this stage, we have an indeterminate form (0/0), which means we cannot directly evaluate the limit. To find the velocity, we need to differentiate the position function with respect to time. The derivative of s(t) = -4.9t^2 + 150 is:
v(t) = -9.8t
Now substitute t = 3 into the velocity function:
v(3) = -9.8(3)
= -29.4 meters per second
Therefore, the velocity of the object when t = 3 seconds is -29.4 meters per second.
b) To find the velocity at which the object impacts the ground, we need to determine the time when the object reaches the ground. We know that the height of the object is given by the position function s(t) = -4.9t^2 + 150.
When the object impacts the ground, its height will be zero. So, we solve the equation:
-4.9t^2 + 150 = 0
To solve this quadratic equation, we can factor it:
t^2 - 150/4.9 = 0
Now, calculate the square root of both sides:
t = ±sqrt(150/4.9)
Calculating the square root gives two values: t ≈ ±5.049
Since time cannot be negative, we consider the positive value:
t ≈ 5.049 seconds
Now, to find the velocity at impact, we substitute t = 5.049 into the velocity function v(t) = -9.8t:
v(5.049) = -9.8(5.049)
= -49.441 meters per second
Therefore, the object will impact the ground with a velocity of approximately -49.441 meters per second.
To find the velocity of the object when t=3, we need to find the limit of the expression as t approaches 3.
a) Find velocity when t=3:
First, substitute t=3 into the given position function:
s(t) = -4.9t^2 + 150
s(3) = -4.9(3)^2 + 150
s(3) = -4.9(9) + 150
s(3) = -44.1 + 150
s(3) = 105.9 meters
Next, substitute t=3 into the expression for velocity:
lim as t approaches 3 (s(3) - s(t)) / (3 - t)
lim as t approaches 3 (105.9 - s(t)) / (3 - t)
Now, plug in t=3 into the velocity expression:
lim as t approaches 3 (105.9 - s(t)) / (3 - t) = lim as t approaches 3 (105.9 - s(3)) / (3 - 3)
lim as t approaches 3 (105.9 - 105.9) / (3 - 3)
lim as t approaches 3 0 / 0
At this point, we get an indeterminate form of 0/0. To evaluate this limit, we can use L'Hopital's Rule, which states that if we have an indeterminate form of 0/0, we can take the derivative of the numerator and denominator separately and then evaluate the limit again.
Taking the derivative of both the numerator and denominator with respect to t:
lim as t approaches 3 (d/dt (105.9 - s(t))) / (d/dt (3 - t))
The derivative of s(t) with respect to t is:
d/dt (-4.9t^2 + 150) = -9.8t
The derivative of (3 - t) with respect to t is:
d/dt (3 - t) = -1
Simplifying the limit expression:
lim as t approaches 3 (-9.8t) / (-1)
= (-9.8 * 3) / (-1)
= -29.4 / -1
= 29.4 meters per second
Therefore, the velocity of the object when t=3 is 29.4 meters per second.
b) To find the velocity at which the object impacts the ground, we need to find the time at which the object reaches the ground. This is the time when s(t) = 0.
Setting s(t) = 0 in the given position function:
-4.9t^2 + 150 = 0
Solving for t:
-4.9t^2 = -150
t^2 = 150 / 4.9
t^2 = 30.6122
t = √(30.6122) or t = -√(30.6122)
Since time cannot be negative, we will only consider the positive root of t:
t = √(30.6122)
t ≈ 5.53 seconds
Now, substitute t = 5.53 into the expression for velocity:
lim as t approaches 5.53 (s(5.53) - s(t)) / (5.53 - t)
lim as t approaches 5.53 (0 - s(t)) / (5.53 - t)
Taking the limit as t approaches 5.53:
lim as t approaches 5.53 (0 - s(t)) / (5.53 - t) = lim as t approaches 5.53 (-s(t)) / (5.53 - 5.53)
lim as t approaches 5.53 (-s(t)) / 0
Again, we get an indeterminate form of -infinity/0. To evaluate this limit, we can once again use L'Hopital's Rule.
Taking the derivative of both the numerator and denominator with respect to t:
lim as t approaches 5.53 (d/dt (-s(t))) / (d/dt (5.53 - t))
The derivative of s(t) with respect to t is:
d/dt (-4.9t^2 + 150) = -9.8t
The derivative of (5.53 - t) with respect to t is:
d/dt (5.53 - t) = -1
Simplifying the limit expression:
lim as t approaches 5.53 (-9.8t) / (-1)
= (-9.8 * 5.53) / (-1)
Now, calculate the value:
≈ 53.894 meters per second
Therefore, the object will impact the ground with a velocity of approximately 53.894 meters per second.