Ben (50 kg) was running to a park from rest. For the first 10 seconds, he reached 6 m/s. He continued to run with constant speed for 20 seconds. Suddenly, he saw an accident and stopped within 10 seconds. You are asked to: (a) find the acceleration for the first 10 seconds (b) calculate his force for running for the first 10 seconds (c) calculate his opposing force for the next 20 seconds (d) find the force needed to make him stop in the last 10 seconds
Answers
Step-by-step explanation:
Let's find the sum of a pair consisting of 10 different digits.
\cdots\longrightarrow0+1+\cdots+8+9=\dfrac{1}{2}\times10\times9⋯⟶0+1+⋯+8+9=
2
1
×10×9
\cdots\longrightarrow0+1+\cdots+8+9=45.⋯⟶0+1+⋯+8+9=45.
Every ten digits repeat 11 times, so the sum is,
\cdots\longrightarrow45\times11=495.⋯⟶45×11=495.
So, the answer is,
\cdots\longrightarrow\boxed{495.}⋯⟶
495.
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Where a,d,la,d,l are the first term, common difference, last term, the arithmetic terms are,
\cdots\longrightarrow a,a+d,a+2d,\cdots,l-2d,l-d,l⋯⟶a,a+d,a+2d,⋯,l−2d,l−d,l
Let S_{n}S
n
express the sum of nn consecutive numbers,
\begin{gathered}\begin{aligned}S_{n}&=a+(a+d)+(a+2d)+\cdots+(l-2d)+(l-d)+l\\S_{n}&=l+(l-d)+(l-2d)+\cdots+(a+2d)+(a+d)+a\end{aligned}\end{gathered}
S
n
S
n
=a+(a+d)+(a+2d)+⋯+(l−2d)+(l−d)+l
=l+(l−d)+(l−2d)+⋯+(a+2d)+(a+d)+a
\cdots\longrightarrow 2S_{n}=n(a+l)⋯⟶2S
n
=n(a+l)
\cdots\longrightarrow\boxed{S_{n}=\dfrac{1}{2}n(a+l).}⋯⟶
S
n
=
2
1
n(a+l).
Where a,n,la,n,l are the first term, number of terms, last term, the arithmetic series is,
\cdots\longrightarrow S_{n}=\dfrac{1}{2}n(a+l)⋯⟶S
n
=
2
1
n(a+l)
\cdots\longrightarrow S_{n}=\dfrac{1}{2}\{a+a+(n-1)d\}⋯⟶S
n
=
2
1
{a+a+(n−1)d}
\cdots\longrightarrow\boxed{S_{n}=\dfrac{1}{2}\{2a+(n-1)d\}.}⋯⟶
S
n
=
2
1
{2a+(n−1)d}.