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Step2247 [10]
2 years ago
11

You will now use Graphical Analysis to calculate the slope of the line. Choose Analyze from the menu. Select Automatic Curve Fit

. Select the proper stock function from the list. (Hint: Your answers from question 1 should help you make the proper selection.) Just below the graph you will see the slope value. Remember that the symbol for slope is "M." When you choose OK to get back to the regular graph screen, you will see a little box on the graph. This box also shows the slope value. (Remember that slope is "m.") If you are unable to use Graphical Analysis, you can also graph your data on paper to determine the slope. Hint: to determine the slope, use the formula m = (y2 - y1) ÷ (x2 - x1). What is the numerical value of the slope of the line? Show your work.
Physics
1 answer:
Vlad1618 [11]2 years ago
3 0
The value for the slope is <span>M=1.13</span>
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Ainat [17]

Answer:

The taken is  t_A  = 19.0 \ s

Explanation:

Frm the question we are told that

  The speed of car A is  v_A  =  22 \ m/s

   The speed of car B is  v_B  = 29.0 \ m/s

     The distance of car B  from A is  d = 300 \ m

     The acceleration of car A is  a_A  = 2.40 \ m/s^2

For A to overtake B

    The distance traveled by car B  =  The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is  

          d = v_B * t_A

Where t_B is the time taken by car B

Now this can also be represented as using equation of motion as

      d = v_A t_A  + \frac{1}{2}a_A t_A^2 - 300

Now substituting values

       d = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

Equating the both d

       v_B * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

substituting values

   29 * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

   7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300

  7 t_A =1.2 t_A^2 - 300

   1.2 t_A^2 - 7 t_A - 300  = 0

Solving this using quadratic formula we have that

     t_A  = 19.0 \ s

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Answer:

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