Anita and Nick are playing tug-of-war near a mud puddle. They are each holding on to an end of a taut rope that has a knot exact
ly in the middle. Anita's position is 6.2 meters east of the center of the puddle and Nick's position is 3.0 meters west of the center of the puddle. What is the location of the knot relative to the center of the puddle? Treat east as positive and west as negative.
1 answer:
Answer:
1.6 meters east
Explanation:
Note: Image attached.
As we can see on the data, Anita is further from the puddle than Nick. Additionally, the distances relatives to the center of the puddle allow us to now the lenght of the rope which is the sum of these two, 9.2 meters. We also know that the knot is at the middle of the rope, so it means it is at 4.6 meters of both Anita and Nick. Now what is next? Well, we now have everything we know. We have the distance between Anita and the puddle and we also know the distance between Anita and the knot (4.6 meters). We just need to subtract the distance to the knot from the distance to the puddle which is equal to 1,6 meters east direction.
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Answer:
a.
b.
c.
Explanation:
Modeling With Functions
Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by
For Carl Lewis's run at the 1987 World Championships, the values of a and b are
Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is
a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?
To compute the accelerations, we must find the function for a as the derivative of v
For t=0
For t=2
b. Find an expression for the distance traveled at time t.
The distance is the integral of the velocity, thus
To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates
Solving for C
Now we complete the equation for the distance
c. Find the time Lewis needed to sprint 100.0 m.
The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.
We are required to find the time at which the distance is 100 m, thus
Rearranging
We define an auxiliary function f(t) to help us find the value of t.
Let's try for t=9 sec
Now with t=9.9 sec
That was a real close guess. One more to be sure for t=10 sec
The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).