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slava [35]
3 years ago
10

A runner covers the last straight stretch of a race in 3 s. During that time, he speeds up from 5 m/s to 11 m/s.What is the runn

er's acceleration in this part of the race? ​
Mathematics
1 answer:
nikitadnepr [17]3 years ago
5 0

Answer:

From first Newton's equation of motion:

{ \bf{v = u + at}}

Substitute the variables:

{ \tt{11 = 5 + (a \times 3)}} \\ { \tt{3a = 6}} \\ { \tt{a = 2 \:  {ms}^{ - 2} }}

Acceleration is 2 ms-²

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\bf \stackrel{\textit{3 lbs of "c"}}{3c}+\stackrel{\textit{5 lbs of "w"}}{5w}~~=~~\stackrel{\textit{costs}}{15} \\\\\\ \stackrel{\textit{12 lbs of "c"}}{12c}+\stackrel{\textit{2 lbs of "w"}}{2w}~~=~~\stackrel{\textit{costs}}{33} \end{cases}\qquad \impliedby \textit{let's use elimination} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{llccccccl} 3c+5w=15&\times (-4)\implies &-12c&+&-20w&=&-60\\ 12c+2w=33&&12c&+&2w&=&33\\ \cline{3-7}\\ &&0&&-18w&=&-27 \end{array}

\bf -18w=-27\implies w=\cfrac{-27}{-18}\implies \blacktriangleright w=\cfrac{3}{2} \blacktriangleleft \\\\\\ \stackrel{\textit{substituting on the 1st equation}}{3c+5\left(\cfrac{3}{2} \right)=15}\implies 3c+\cfrac{15}{2}=15 \implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{2}}{2\left( 3c+\cfrac{15}{2} \right)=2(15)} \\\\\\ 6c+15=30\implies 6c=15\implies c=\cfrac{15}{6}\implies \blacktriangleright c=\cfrac{5}{2} \blacktriangleleft

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