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nikklg [1K]
3 years ago
8

Skylar and Rodrigo each recorded how far they traveled while skateboarding. Skylar traveled 65 feet in 5 seconds and Rodrigo tra

veled 108 feet in 8 seconds. How much farther did Rodrigo travel per second than Skylar?
Mathematics
1 answer:
nikklg [1K]3 years ago
7 0

9514 1404 393

Answer:

  1/2 ft

Step-by-step explanation:

Skylar's speed was ...

  (65 ft)/(5 s) = 13 ft/s

Rodrigo's speed was ...

  (108 ft)/(8 s) = 13.5 ft/s

Rodrigo traveled farther by (13.5 -13) = 0.5 ft per second.

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the easiest approach with a given point and the slope of the line is the point-slope form :

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jolli1 [7]

Answer:

the answer is incomplete, below is the complete question

"Reparametrize the curve with respect to arc length measured from the point where t = 0 in the direction of increasing t. (Enter your answer in terms of s.) r(t) = 3ti + (1 - 4t)j + (1 + 2t)k r(t(s)) ="

answer

r(t(s)) = \frac{3s}{\sqrt{29} } i + (1 -\frac{4s}{\sqrt{29} }t)j + (1 + \frac{2s}{\sqrt{29} })k

Step-by-step explanation:

The step by step procedure is to first determine the differentiate the given vector function

r(t) = 3ti + (1 - 4t)j + (1 + 2t)k

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s(t)=\int\limits^t_0 {||r'(t)||} \, dt

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s(t)=\int\limits^t_0 {||r'(t)||} \, dt\\s(t)=\int\limits^t_0 {\sqrt{3^{2} +4^{2}+2^{2}} \, dt\\s(t)=\int\limits^t_ 0{\sqrt{29} } \, dx\\

s(t)=\int\limits^t_ 0{\sqrt{29} } \, dx\\\\s(t)=\sqrt{29} t\\hence \\t(s)=\frac{s}{\sqrt{29} }

substituting the value of t in to the given vector equation we have

r(t(s)) = \frac{3s}{\sqrt{29} } i + (1 -\frac{4s}{\sqrt{29} }t)j + (1 + \frac{2s}{\sqrt{29} })k

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