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Harlamova29_29 [7]
3 years ago
12

(a) The position of a particle at time t is s(t) = t^3 + t. Compute the average velocity over the time interval [7, 12].(b) Esti

mate the instantaneous velocity at t = 7. (Round your answer to the nearest whole number)
Mathematics
1 answer:
ryzh [129]3 years ago
8 0

Answer:

<h2>a) Average velocity = 278 units</h2><h2>b) Instantaneous velocity at t = 7 seconds is 148 units</h2>

Step-by-step explanation:

a) Average velocity is the ratio of displacement to time.

   We have

              s(t) = t³ + t

    t is in between 7 and 12

             s(7) =  7³ + 7 = 350

              s(12) = 12³ + 12 = 1740

    Displacement = 1740 - 350 = 1390

   Time = 12 - 7 = 5

  Displacement =  Average velocity x time

  1390 = Average velocity x 5

  Average velocity = 278 units

b) s(t) = t³ + t

   Differentiating

         v(t) = 3t² + 1

  At t = 7

          v(t) = 3 x 7² + 1 = 148 units

  Instantaneous velocity at t = 7 seconds is 148 units

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Note that powers of 2 can be written in binary as

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Also observe that

\log_2(2^n)=n\log_22=n

so we need only add 1 to the logarithm to find the number of binary digits needed to represent powers of 2. For any other number (non-power-of-2), we would need to round down the logarithm to the nearest integer, since for example,

2_{10}=10_2\iff\log_2(2^1)=\log_22=1
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Now, 2^9=512, and 999 falls between these consecutive powers of 2. That means

\log_2999=9+\text{(some number between 0 and 1})

which means 999 requires \lfloor\log_2999\rfloor+1=9+1=10 binary digits.

Your question seems to ask how many binary digits in total you need to represent all of the numbers 0-999. That would depend on how you encode numbers that requires less than 10 digits, like 1. Do you simply write 1_2? Or do you pad this number with 0s to get 10 digits, i.e. 0000000001_2? In the latter case, the answer is obvious; 1000\times10=10^4 total binary digits are needed.

In the latter case, there's a bit more work involved, but really it's just a matter of finding how many number lie between successive powers of 2. For instance, 0 and 1 both require one digit, 2 and 3 require two, while 4-7 require three, while 8-15 require four, and so on.
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Answer:

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