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Updated on: 12 Jun 2024, 01:20
Originally posted by imhimanshu on 16 Sep 2012, 00:25.
Last edited by Bunuel on 12 Jun 2024, 01:20, edited 9 times in total.
Last edited by Bunuel on 12 Jun 2024, 01:20, edited 9 times in total.
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\(v_{0} \): 5
\(v_{10} \): 20
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
67% (02:30) correct
33%
(02:57)
wrong
based on 3075
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A car is traveling on a straight stretch of roadway, and the speed of the car is increasing at a constant rate with respect to time. At time 0 seconds, the speed of the car is \(v_0\) meters per second; 10 seconds later, the front bumper of the car has traveled 125 meters, and the speed of the car is \(v_{10}\) meters per second.
Select values of \(v_0\) and \(v_{10}\) that are together consistent with the information provided. Make only two selections, one in each column.
Select values of \(v_0\) and \(v_{10}\) that are together consistent with the information provided. Make only two selections, one in each column.
ID: 100401
| \(v_{0} \) | \(v_{10} \) | |
| 5 | ||
| 18 | ||
| 20 | ||
| 36 | ||
| 72 |
ShowHide Answer
Official Answer
\(v_{0} \): 5
\(v_{10} \): 20
17 Sep 2012, 11:53
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imhimanshu
Responding to a pm:
Your best friend in GMAT is your reasoning skill. You can solve this question easily using brute force. Of course there is nothing wrong with the complete AP approach but I would use it only partially...
The car travels 125 m in 10 sec. It's average speed is 12.5 m/s
vo can only be 5. If it were 18, the car would have traveled more than 180m in 10 sec since the minimum speed is 18. The speed keeps increasing thereafter. Same logic goes for all other values.
Now, with vo = 5, which value will give the average as 12.5? Obviously 20. (recall that average of an AP is the average of its first and last terms)
23 Oct 2013, 11:56
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danzig
Dear Danzig & Mbmanoj & Chakdum,
This is tricky. When a car is increasing at constant acceleration, then it is NOT a sequence, either arithmetic or geometric. Thinking about the motion in terms of a sum of what happens in each second is not helpful, and in particular, thinking of it as a sum of constant-motion chunks each second is DEAD WRONG. If it goes from, say 5 m/s to 15 m/s in 10 seconds, then the acceleration is 2 m/s^2, but that does not mean: 5 m/s for the duration of the 1st second, 7 m/s for the duration of the 2nd second, etc. That is a complete misunderstanding of the nature of acceleration.
Instead, the formula given in the OE, average velocity = (vo + vf)/2, is always correct for constant acceleration. Here's one way to think about that formula. Think about the graph of speed vs. time. The speed is continuously increasing from (vo) to (vf).
Attachment:
constant acceleration v vs. t.JPG [ 16.64 KiB | Viewed 44670 times ]
Area of a Trapezoid = (average of the parallel bases)*(height)
Here, the parallel bases are the two vertical segments --- the one on the left has a length of (vo) and the one on the right has a length of (vf), so we average those two. The trapezoid is flipped on its edge, so the "height" (i.e. the distance between the two parallel segments) is the horizontal length at the bottom --- 10 s. Thus
Area = (10 s)*(vo + vf)/2 = distance traveled
Now, think about average velocity (AV). We know that this relates total distance (DT) and total time (TT) of any trip.
(DT) = (AV)*(TT)
DT = 10*(AV)
But from the equation above, from the area of the trapezoid, we know
DT = 10*(vo + vf)/2
Comparing those two makes immediately clear:
AV = (vo + vf)/2
This formula has nothing to do with a sequence of any kind. It comes from the area of a trapezoid! For this problem, it is a very useful shortcut:
DT = 125
TT = 10
125 = 10*(vo + vf)/2
125 = 5*(vo + vf)
25 = (vo + vf)
So we just need to numbers that have a sum of 25.
Does all this make sense?
Mike
