An eagle is flying around and its velocity vvvv as a function of time tttt is given in the graph below where rightwards is the p
ositive velocity direction.
A set of black coordinate axes are given with the vertical axis labeled "v (m/s)" and the horizontal axes labeled "t (s)". A curve that relates v to t is shown in blue. It begins with a straight line of endpoints (0,-2) and (3,4). This first line is connected to a second line with endpoints (3,4) and (4,4).
A set of black coordinate axes are given with the vertical axis labeled "v (m/s)" and the horizontal axes labeled "t (s)". A curve that relates v to t is shown in blue. It begins with a straight line of endpoints (0,-2) and (3,4). This first line is connected to a second line with endpoints (3,4) and (4,4).
What is the eagle's displacement Δx\Delta xΔxdelta, x from t=1.0 st=1.0\,\text st=1.0st, equals, 1, point, 0, start text, s, end text to 4.0 s4.0\,\text s4.0s4, point, 0, start text, s, end text?
A set of black coordinate axes are given with the vertical axis labeled "v (m/s)" and the horizontal axes labeled "t (s)". A curve that relates v to t is shown in blue. It begins with a straight line of endpoints (0,-2) and (3,4). This first line is connected to a second line with endpoints (3,4) and (4,4).
A set of black coordinate axes are given with the vertical axis labeled "v (m/s)" and the horizontal axes labeled "t (s)". A curve that relates v to t is shown in blue. It begins with a straight line of endpoints (0,-2) and (3,4). This first line is connected to a second line with endpoints (3,4) and (4,4).
What is the eagle's displacement Δx\Delta xΔxdelta, x from t=1.0 st=1.0\,\text st=1.0st, equals, 1, point, 0, start text, s, end text to 4.0 s4.0\,\text s4.0s4, point, 0, start text, s, end text?
1 answer:
9514 1404 393
Answer:
8 units
Step-by-step explanation:
Displacement is the area under the velocity curve. The area of interest is in the shape of a trapezoid, so we can use the formula ...
A = (1/2)(b1 +b2)h
to find the area.
Here, b1=3, b2=1, h=4, so the area is ...
A = (1/2)(3 +1)(4) = 8
The displacement is 8 units in the time from t=1 to t=4.
You might be interested in
When two lines intersect, the angles across from the intersection (the "X") are congruent (equal), and referred to as "vertical angles."
So by the rule of vertical angles are congruent, we can set them equal:
to factor we need the factors of 84:
1×84, 2×42, 3×28, 4×21, 6×14
Now we need one of those factors ×2 ( from the x^2 term) minus the other factor to equal 13 (the middle term)
it turns out that if we use 4×21: 4×2 = 8
and 21-8 = 13. Woohoo! we got it
So now we have:
Now set each factor = 0
1) 2x - 21 = 0 and 2) x + 4 = 0
1) 2x - 21 = 0, 2x = 21, x = 21/2 = 10.5
2) x + 4 = 0, x = -4
Now we plug in each answer to check:
1) x = 10.5
1/2x + 42 = 1/2(10.5) + 42 = 5.25 + 42 = 47.25
47.25 = 47.25
BAM SO IT LOOKS LIKE THAT'S IT!!
But let's check 2) first:
2) x = -4
1/2x + 42 = 1/2(-4) + 42 = -2 + 42 = 40
40 = 40, so both work!!
But it asks for m<1, and <1 is supplementary because it lies on same line, meaning the sum of both = 189
Therefore m<1 = 180 - 47.25 = 132.75°
OR m<1 = 180 - 40 = 140°
So by the rule of vertical angles are congruent, we can set them equal:
to factor we need the factors of 84:
1×84, 2×42, 3×28, 4×21, 6×14
Now we need one of those factors ×2 ( from the x^2 term) minus the other factor to equal 13 (the middle term)
it turns out that if we use 4×21: 4×2 = 8
and 21-8 = 13. Woohoo! we got it
So now we have:
Now set each factor = 0
1) 2x - 21 = 0 and 2) x + 4 = 0
1) 2x - 21 = 0, 2x = 21, x = 21/2 = 10.5
2) x + 4 = 0, x = -4
Now we plug in each answer to check:
1) x = 10.5
1/2x + 42 = 1/2(10.5) + 42 = 5.25 + 42 = 47.25
47.25 = 47.25
BAM SO IT LOOKS LIKE THAT'S IT!!
But let's check 2) first:
2) x = -4
1/2x + 42 = 1/2(-4) + 42 = -2 + 42 = 40
40 = 40, so both work!!
But it asks for m<1, and <1 is supplementary because it lies on same line, meaning the sum of both = 189
Therefore m<1 = 180 - 47.25 = 132.75°
OR m<1 = 180 - 40 = 140°