The x-coordinates of two objects moving along the x-axis are given as a function of time t. x1 = (4 m/s)t and x2 = −(159 m) + (2
4 m/s)t − (1 m/s2)t2 . Calculate the magnitude of the distance of closest approach of the two objects. x1 and x2 never have the same value.
1 answer:
The x-coordinate of the first object is
x₁(t) = (4 m/s)t
The x-coordinate of the second object is
x₂(t) = -(159 m) + (24 m/s)t - (1 m/s²)t²
The distance between the two objects is
x(t) = x₂ - x₁
= - 159 + 24t - t² - 4t
= -t² + 20t - 159
Write this equation in the standard form for a parabola.
x = -[t² - 20t] - 159
= -[(t - 10)² - 100] - 159
= -(t-10)² - 59
This parabola has a vertex at (-10, -59), and it is downward.
Because the maximum value of x is negative, the two objects never touch
The closest distance between the objects is 59 m.
The two graphs confirm that the analysis is correct.
Answer: The closest approach is 59 m.
x₁(t) = (4 m/s)t
The x-coordinate of the second object is
x₂(t) = -(159 m) + (24 m/s)t - (1 m/s²)t²
The distance between the two objects is
x(t) = x₂ - x₁
= - 159 + 24t - t² - 4t
= -t² + 20t - 159
Write this equation in the standard form for a parabola.
x = -[t² - 20t] - 159
= -[(t - 10)² - 100] - 159
= -(t-10)² - 59
This parabola has a vertex at (-10, -59), and it is downward.
Because the maximum value of x is negative, the two objects never touch
The closest distance between the objects is 59 m.
The two graphs confirm that the analysis is correct.
Answer: The closest approach is 59 m.
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Answer:
Explanation:
For an object to move with constant velocity, the acceleration of the object must be zero:
.
As the net force equals acceleration multiplied by mass , this must mean:
.
So, the sum of the three forces must be zero:
,
this implies:
.
To obtain this sum, its easier to work in Cartesian representation.
First we need to define an Frame of reference. Lets put the x axis unit vector pointing east, with the y axis unit vector
pointing south, so the positive angle is south of east. For this, we got for the first force:
,
as is pointing north, and for the second force:
,
as is pointing west.
Now, our third force will be:
But, we need the magnitude and the direction.
To find the magnitude, we can use the Pythagorean theorem.
this is the magnitude.
To find the direction, we can use:
and this is the angle south of east.