Related questions with answers

The frequency table summarizes the distribution of time that 140 patients at the emergency room of a small-city hospital waited to receive medical attention during the last month.

 Waiting time  Frequency  Less than 10 minutes 5 At least 10 but less than 20 minutes 24 At least 20 but less than 30 minutes 45 At least 30 but less than 40 minutes 38 At least 40 but less than 50 minutes 19 At least 50 but less than 60 minutes 7 At least 60 but less than 70 minutes 2\begin{array}{lr} \hline \text { Waiting time } & \text { Frequency } \\ \hline \text { Less than } 10 \text { minutes } & 5 \\ \text { At least } 10 \text { but less than } 20 \text { minutes } & 24 \\ \text { At least } 20 \text { but less than } 30 \text { minutes } & 45 \\ \text { At least } 30 \text { but less than } 40 \text { minutes } & 38 \\ \text { At least } 40 \text { but less than } 50 \text { minutes } & 19 \\ \text { At least } 50 \text { but less than } 60 \text { minutes } & 7 \\ \text { At least } 60 \text { but less than } 70 \text { minutes } & 2 \\ \hline \end{array}

Which of the following represents possible values for the median and IQR of waiting times for the emergency room last month? (a) median = 27 minutes and IQR = 15 minutes (b) median = 28 minutes and IQR = 25 minutes (c) median = 31 minutes and IQR = 35 minutes (d) median = 35 minutes and IQR = 45 minutes (e) median = 45 minutes and IQR = 55 minutes

Question

(a) A light-rail commuter train accelerates at a rate of 1.35 m/s². How long does it take to reach its top speed of 80.0 km/h, starting from rest? (b) The same train ordinarily decelerates at a rate of 1.65 m/s². How long does it take to come to a stop from its top speed? (c) In emergencies, the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency acceleration in meters per second squared?

Solution

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(a) Train starts from rest. So, v0=0m.s1v_0 = 0\:m.s^{-1}

Its top speed is v=80.0kmh=80.0×103m3600s=22.22m.s1v=80.0\:\dfrac{km}{h} = 80.0 \times \dfrac{10^3\:m}{3600\:s} = 22.22\:m.s^{-1}

where 1km=103m1\:km = 10^3\:m and 1h=3600s1\:h = 3600\:s

Acceleration of train is a=1.35m.s2a=1.35\:m.s^{-2}

We have the following kinematic equation,

v=v0+atv = v_0 + at

Substituting the given values

22.22=0+1.35t22.22 = 0 + 1.35 t

t=22.221.3516.5st = \dfrac{22.22}{1.35} \:\approx\:16.5\:s

So the time taken to reach the top speed from rest at the given acceleration is 16.5s16.5\:s.

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