M09 #31

Actual Distance = Velocity x 1 second

(1) \(D = \frac{5^2}{100} + 9 \hspace{10} \Rightarrow Distance_{safe} \hspace{5} = \hspace{5} 9.25 \hspace{5} > \hspace{5} Distance_{actual} \hspace{3}(5 m)\) BREACH

(2) \(D = \frac{10^2}{100} + 9 \hspace{10} \Rightarrow Distance_{safe} \hspace{5} = \hspace{5} 10 \hspace{5} = \hspace{5} Distance_{actual} \hspace{3}(10 m)\) On Par, FINE

(3) \(D = \frac{25^2}{100} + 9 \hspace{10} \Rightarrow Distance_{safe} \hspace{5} = \hspace{5} 15.25 \hspace{5} < \hspace{5} Distance_{actual} \hspace{3}(25 m)\) FINE

(4) \(D = \frac{50^2}{100} + 9 \hspace{10} \Rightarrow Distance_{safe} \hspace{5} = \hspace{5} 34 \hspace{5} < \hspace{5} Distance_{actual} \hspace{3}(34 m)\) FINE

(5) \(D = \frac{90^2}{100} + 9 \hspace{10} \Rightarrow Distance_{safe} \hspace{5} = \hspace{5} 90 \hspace{5} = \hspace{5} Distance_{actual} \hspace{3}(90 m)\) On Par, FINE


Clearly, first choice of 5 m/s speed violates the regulation because distance maintained (5 m) is less than what the rule suggests.

Answer A.

I think you got confused which distance should be greater than the other.

I understand that plugging values is a better way but this is how i did it....

"If two cars want to maintain a time interval of 1 second between each other"
think of it this way => distant is given interms of time (1 sec)..ie ...'D' meters / 1 sec (or D m/s) hence velocity to be maintained = distance => substitute D=V or V=D in above equation => we get V^2 - 100V + 900 < 0 ( < sign indicates required condition ..ie..Speed which will not breach the safety barrier ) => on solving this we get (V-90) (V-10) < 0 => thus it is clear that any speed between 10 m/s to 90 m/s is OK but nothing out of this range....Hence answer is 5 m/s

If you like my post...consider it for Kudos...

Great explanation! Thank you!

Yeah punching the numbers in for the velocity .This explation makes it very clear for option A to be correct.

I am getting confused with the equation

on LHS we have a variable (D) whose unit is m whereas on RHS we have (v^2) which is (m/s)^2... so how is the equation valid??
Please explain

One more Quick way to do.
Looking at the equation...I choose the most easy value for plugging in..i.e., 10 m/s .

D = 100/100 + 9 = 10, which exactly meeting the norms (Option B).

Now, the fun part is you have
3 options (C,D,E) > just Ok value (10)
1 option (B) < just ok value (10).

I didn't use my brain much analyzing which is exceeding or less than norms...
All I know is there should be ONE & ONLY answer...hence option "A".


Gimme a kudo if you like this

I did using the bhushan252 way.......but got confused in the end.....niw i understand it.......A

A

Not sure if I did this right but here goes:

V= d/t therefore V = d/1 therefore V=D
D = (D^2/100) + 9
100D = D^2 + 900
D^2 -100D + 900
D = 10 or 90
Therefore, to meet the specifications of the formula the answer must be between 10 and 90 so the answer must be A.

I have no idea if I did it right, but I did get the right answer!

Here is how I solved it:

T = D/R; (v^2+900)/100r = 1. v = 90, v = 10. Since the distance should be less, "A" must be correct.

Hope somebody gets this strange approach.

Nice explanation. Got confused too.

tough one to understand...

for safe distance -

D >= V^2/100 + 9



D = V*T = V (T=1, given)

D >= D^2/100 + 9

D^2 -100D +9 =< 0

(D-10)(D-90) =< 0

10 =< D =< 90
10 =< V =< 90 (V=D)
anything, out of this range would violate the speed

only option A violates the speed limit

The biggest issue is the awkward wording of the question. It would be better if it was clarified that breach meant go in excess of as opposed to break. Since you can in theory breach it by going to close or too far. Since Only B and E work out to be exact it should be reworded as such that going over the amount is clear that its incorrect.