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.Sam.
 4 years ago
In order that a train can stop safely, it will always pass a signal showing a yellow light before it reaches a signal showing a red light. Drivers apply the brake at the yellow light and this results in a uniform deceleration to stop exactly at the red light.
The distance between the red and yellow lights is x.
What must be the minimum distance between the lights if the train speed is increased by 20 %, without changing the deceleration of the trains?
.Sam.
 4 years ago
In order that a train can stop safely, it will always pass a signal showing a yellow light before it reaches a signal showing a red light. Drivers apply the brake at the yellow light and this results in a uniform deceleration to stop exactly at the red light. The distance between the red and yellow lights is x. What must be the minimum distance between the lights if the train speed is increased by 20 %, without changing the deceleration of the trains?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the minimum distace b/w yellwo and red light should be increas by 6%.

stormfire1
 4 years ago
Best ResponseYou've already chosen the best response.1Since you have a constant acceleration (which is negative in this instance) you can start with this kinematic equation: \[V_f^2=V_i^2+2a(\Delta x)\]Since Vf = 0 m/s the equation can be rewritten as: \[V_i^2+2a(\Delta x)=0\]Solving this equation for delta x gives you:\[\Delta x=\frac{Vi^2}{2a}\](Note that since acceleration is negative so you will end up with a postive delta x value) Now that you have this equation you should immediately see that if you increase Vi by some value, x will be increased by the square of that value. In your problem, you end up with: \[1.2^2 \Delta x=1.44 \Delta x=\frac{1.2Vi^2}{2a}\]so the original stopping distance (delta x) must be increased by 44%.

stormfire1
 4 years ago
Best ResponseYou've already chosen the best response.120% increase in velocity is the same as 1.2*velocity
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