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Use quadratic functions for modelling motion

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A new set of automatic sliding doors at the entrance to a supermarket is being designed. The doors will consist of a pair of 100cm wide glass panels which are programmed to slide open in opposite directions when a sensor is triggered. The panels are identical except for the direction in which they move. For the purposes of this question you will consider the motion of the right-hand panel, Let y be the position, measured in centimetres, of the left edge of the panel relative to its position when closed. The position is modelled as a function of time, t, in secondes.

(i) It proposed that the doors should be programmed to open in two stages with the motion of the glass panel described by the following two equations.

stage position time period

1 50t 0 < t < 1
2 y= -50t +200t-100 1 < t < 2
- -

(a) For each stage of motion plot the function on your calculator, using an appropriate window foir the given time interval. Hence sketch the position-time graph over the range 0 to 2 seconds, adding appropriate labelling ans indicating each stage of motion clearly.

(b) for each stage of motion, plot the rate of change of the function. Hence sketch the velocity - time graph over the range 0 to 2 seconds labelling it clearly

(c) Copy and complete table 2, by finding the position, velocity and acceleration of the panel at the tines listed.

Table 2

Stage time/ position/ Velocity/ Acceleration
seconds cm cm s -1 cm s-1

1 0
1 0.5
2 1.5
2 2

(ii) After some discussion, the managers of the supermarket decide that the sliding doors should be programmed differently. They would like the motion of the panels to be described by an equation of the form
y= At +Bt +C for 0 < t< 2. The following features are
_ _
also required: at time t=0 the position should be y=0, and the parabola described by the equation should have a vertex at(t,y) = (2,100).

(a) Find the values of A, B and C that satisfy these requirements.

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<br>a) From the question you provided, i think the motion of the glass panel described by the following two equations.
<br> 1). y=50t. 0<t<1.
<br> 2). y=-50t^2+200t-100=-50(t-2)^2+100, 1<t<2.
<br>Note. in your original stating, you gave y=-50t+200t-100 for 1<t<2.
<br>I guessed that you should give a equation as above. If not, the result will be wrong. I give an answer related the equation 1) and 2).
<br> The graph for 0<t<1 is a line through origin O(0,0). The senond part for 1<t<2 is a parabola with ...

Solution provided by:
  • BSc , Wuhan Univ. China
  • MA, Shandong Univ.
Recent Feedback
  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
  • "excellent work"
  • "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
  • "Thank you"
  • "Thank you very much for your valuable time and assistance!"
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