anonymous
  • anonymous
The car velocity is 100 km/h. Suddenly, the driver sees a elk 120 m from the car standing on the road. The driver’s reaction time is 1,5 s, after which time the driver starts braking. The braking is assumed steady and the deceleration is 7,5 m/s2. The car stops in 4 s. Calculate the distance between the standing elk and the car after the car has stopped. Give the answer in 1 m precision.
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
Okay...I'm going to type this out in a few parts...
anonymous
  • anonymous
Firstly, are you studying kinematics?
anonymous
  • anonymous
The following equation gives you the distance traveled by an object under an acceleration a, initial velocity v_0 and initial displacement x_0:\[x=\frac{1}{2}at^2+v_0t+x_0\]

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anonymous
  • anonymous
I am not studying kinematics
anonymous
  • anonymous
Now, the driver is traveling at 100km/h which equals 250/9 m/s. It takes the driver 1.5 seconds before any acceleration takes place, so they will travel an initial distance at first of \[(distance)=(speed)(time)=\frac{250}{9}m/s \times 1.5 s=\frac{125}{3}m\]
anonymous
  • anonymous
So the distance the car travels in total will be\[x=\frac{1}{2}(-7.5)m/s^2 \times (4s)^2+\frac{250}{9}m/s \times 4 s+\frac{125}{3}m\]\[=\frac{835}{9}m \approx92.78m\]
anonymous
  • anonymous
i am finaly starting to understand it, you are really good at explaining, are you a sicience lecturer?
anonymous
  • anonymous
But this is the distance traveled by the car. The distance between the elk and the car (the distance left over) is 120m - (825/9)m = 245/9m; that is\[x=\frac{245}{9}m \approx 27.22m\] which is 27m (to nearest metre).
anonymous
  • anonymous
I'm a graduate in mathematics and physics, and teach it to secondary and tertiary students to pay for my postgraduate studies.
anonymous
  • anonymous
wow, nice thanx for helping
anonymous
  • anonymous
No probs. If you haven't 'fanned' me, I'd appreciate it - I want to get my points up.
anonymous
  • anonymous
i already have

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