Answer:b
Step-by-step explanation:
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
Answer:
The answer is 5.75
Step-by-step explanation:
I used desmos
840 centimeters ...there 100 centimeters in one meter
The slide needs to be 4.123 meters
Step-by-step explanation:
Length for the ladder- 1 mtr
Distance of the bottom of the slide from the ladder- 4 mtr
Length of the slide needed
We can view the whole set up in the form of a right-angled triangle with the base representing the slide distance from the ladder, ladder length representing altitude and slide length representing as hypotenuse (As provided in the diagram)
For right-angled triangle we know from Pythagoras theorem-
Base² + altitude²= hypotenuse
²
Substituting the values of base as 4 mtr and altitude as 1 m
(Hypotenuse)²= 17
∴ Hypotenuse (slide length) =4.123 mtr
∴ Length of the slide needed= 4.123 meters