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A car is travelling on a straight stretch of raodway

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A car is travelling on a straight stretch of raodway [#permalink]

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16 Sep 2012, 00:25
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Could you please solve the below attached question
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Re: A car is travelling on a straight stretch of raodway [#permalink]

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16 Sep 2012, 01:30
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imhimanshu wrote:
Could you please solve the below attached question

Wow it looks like a different Question. I am trying to solve it with a physics average speed formula: Avg Velocity : (u+v)/2, where u=initial velocity, V=final velocity.

From Question We understand the Car travels 125 m in 10 sec. Therefore the average velocity = 125/10 m/s or 12.5 or 25/2.

Now from the formula (u+v)/2 = 25/2, Hence u+v=25. By scanning the answer choices answer choice u=5 & v=20 fits. Hence it may be the answer.
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Re: A car is travelling on a straight stretch of raodway [#permalink]

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16 Sep 2012, 12:42
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this question is a application of AP. and yes the previous answers are correct.

Speed at time '0' = v0
Speed at time '10' = v10
Lets assume the speed is increasing at a rate of 'd' m/s
Note:- Distance traveled in any 1 second is equal to the speed during that second.
Distance traveled in 1st sec = v0
Distance traveled in 2nd sec = v0+d
Distance traveled in 3rd sec = v0+2d
Distance traveled in 10th sec = v0+9d

So Total Distance traveled in 10 second can be given by
v0 + (v0+d) + (v0+2d)+.........+(v0+8d)+(v0+9d) = 10/2[2v0 + (10-1)d]=10/2[2v0 + 9d]

& as per question the bumper has traveled 125 m at the end of 10th second. So
10/2[2v0 + 9d]= 125
[2v0 + 9d]= 25
v0 + (v0+9d) = 25
v0 + v10 = 25......(1)

We are asked the value for speed at 0th & 10th second. From equation (1) we can say that the sum of speed at these moments is equal to 25.
The only options available are 5 , 20.

The speed v0 must be 5 & v10 must 20 because the speed is increasing at constant rate.
Note:- If the speed was decreasing at a constant rate, then v0=20 & v10=5

Hope it helps.
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Re: A car is travelling on a straight stretch of raodway [#permalink]

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17 Sep 2012, 09:08
SOURH7WK wrote:
imhimanshu wrote:
Could you please solve the below attached question

Wow it looks like a different Question. I am trying to solve it with a physics average speed formula: Avg Velocity : (u+v)/2, where u=initial velocity, V=final velocity.

From Question We understand the Car travels 125 m in 10 sec. Therefore the average velocity = 125/10 m/s or 12.5 or 25/2.

Now from the formula (u+v)/2 = 25/2, Hence u+v=25. By scanning the answer choices answer choice u=5 & v=20 fits. Hence it may be the answer.

In this case, your reasoning is correct. The average speed is the mean between the initial and the final speed because the sequence of the speeds for every minute is an arithmetic progression (each speed is the previous one increased by the same amount).
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Re: A car is travelling on a straight stretch of raodway [#permalink]

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17 Sep 2012, 09:24
Wow this appears to be more of a physics question :D

Can be solved by
s =(v+u)*t /2

or combining
v=u+at and
v^2 =u^2+2as

imhimanshu wrote:
Could you please solve the below attached question

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Re: A car is travelling on a straight stretch of raodway [#permalink]

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17 Sep 2012, 11:53
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imhimanshu wrote:
Could you please solve the below attached question

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)
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Re: A car is travelling on a straight stretch of raodway   [#permalink] 17 Sep 2012, 11:53
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