You can solve for the velocity and position functions by integrating using the fundamental theorem of calculus:
<em>a(t)</em> = 40 ft/s²
<em>v(t)</em> = <em>v </em>(0) + ∫₀ᵗ <em>a(u)</em> d<em>u</em>
<em>v(t)</em> = -20 ft/s + ∫₀ᵗ (40 ft/s²) d<em>u</em>
<em>v(t)</em> = -20 ft/s + (40 ft/s²) <em>t</em>
<em />
<em>s(t)</em> = <em>s </em>(0) + ∫₀ᵗ <em>v(u)</em> d<em>u</em>
<em>s(t)</em> = 10 ft + ∫₀ᵗ (-20 ft/s + (40 ft/s²) <em>u</em> ) d<em>u</em>
<em>s(t)</em> = 10 ft + (-20 ft/s) <em>t</em> + 1/2 (40 ft/s²) <em>t</em> ²
<em>s(t)</em> = 10 ft - (20 ft/s) <em>t</em> + (20 ft/s²) <em>t</em> ²
Y = 1/3x - 2.
If you sub in 0, you’ll get (0, -2) which is the y intercept.
Answer:
ANSWER: (4, -12)
Step-by-step explanation:
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