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anonymous
 5 years ago
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.
anonymous
 5 years ago
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.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay...I'm going to type this out in a few parts...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Firstly, are you studying kinematics?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The following equation gives you the distance traveled by an object under an acceleration a, initial velocity v_0 and initial displacement x_0:\[x=\frac{1}{2}at^2+v_0t+x_0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am not studying kinematics

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now, 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
 5 years ago
Best ResponseYou've already chosen the best response.0So 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
 5 years ago
Best ResponseYou've already chosen the best response.0i am finaly starting to understand it, you are really good at explaining, are you a sicience lecturer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But 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
 5 years ago
Best ResponseYou've already chosen the best response.0I'm a graduate in mathematics and physics, and teach it to secondary and tertiary students to pay for my postgraduate studies.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wow, nice thanx for helping

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No probs. If you haven't 'fanned' me, I'd appreciate it  I want to get my points up.
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