(a) By the fundamental theorem of calculus,
<em>v(t)</em> = <em>v(0)</em> + ∫₀ᵗ <em>a(u)</em> d<em>u</em>
The particle starts at rest, so <em>v(0)</em> = 0. Computing the integral gives
<em>v(t)</em> = [2/3 <em>u</em> ³ + 2<em>u</em> ²]₀ᵗ = 2/3 <em>t</em> ³ + 2<em>t</em> ²
(b) Use the FTC again, but this time you want the distance, which means you need to integrate the <u>speed</u> of the particle, i.e. the absolute value of <em>v(t)</em>. Fortunately, for <em>t</em> ≥ 0, we have <em>v(t)</em> ≥ 0 and |<em>v(t)</em> | = <em>v(t)</em>, so speed is governed by the same function. Taking the starting point to be the origin, after 8 seconds the particle travels a distance of
∫₀⁸ <em>v(u)</em> d<em>u</em> = ∫₀⁸ (2/3 <em>u</em> ³ + 2<em>u</em> ²) d<em>u</em> = [1/6 <em>u</em> ⁴ + 2/3 <em>u</em> ³]₀⁸ = 1024
The equation of the line graphed below would be D. I can find two points that are easily distinguishable on this graph. (1, 5) and (0, 0). I found the slope by subtracting 0 from 5 giving me -5. Next, I subtracted 0 and 1 giving me -1. This gives me a fraction of -5/-1 which equals to 5.
Answer:
52x + 26y + 20z
Step-by-step explanation:
Assuming you needed the answer to be simplified!
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Both graphs A and B are linear. In both graphs, the y value is going up at a constant rate in comparison to their x values. For example, in graph A every time the x value is increased by 1 the y value is increased by 2 values showing that they have the same gradient. For a graph to be linear it must have a constant rate of increase or decrease.
