Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 29 May 2016, 07:10

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# IR - speed

Author Message
Manager
Joined: 11 Aug 2012
Posts: 128
Schools: HBS '16, Stanford '16
Followers: 1

Kudos [?]: 72 [1] , given: 16

### Show Tags

22 Jul 2013, 15:56
1
KUDOS
A car is traveling on a straight stretch of roadway, and the speed of the car is increasing at a constant rate. At time 0 seconds, the speed of the car is V0 meters per second; 10 seconds later, the front bumper of the car has traveled 125 meters and the speed of the car is V10 meters per second.

In the table below, select values of V0 and V10 that are together consistent with the information provided. Make only two selections, one in each column.

My doubt is:
The question indicates that the speed of the car is increasing at a constant rate. So, is it talking about an arithmetic sequence, or a geometric sequence? According to the OE, the average speed is $$\frac{1}{2}*( V0 + V10)$$. So, it seems that it is talking about an arithmetic sequence because that's the way we use to calculate the average in an arithmetic sequence. Please confirm.
However, I remember that, when a question mentions that something is increasing at a constant rate, we must multiply the first value by a constant, we shouldn't add. Please, your help.

OA:
[Reveal] Spoiler:
V0 = 5 ; V10 = 20

Attachments

44.png [ 3.05 KiB | Viewed 2864 times ]

Intern
Joined: 06 Sep 2012
Posts: 14
Concentration: Entrepreneurship, Sustainability
GPA: 3.11
Followers: 0

Kudos [?]: 4 [0], given: 2

### Show Tags

31 Jul 2013, 09:52
danzig wrote:
A car is traveling on a straight stretch of roadway, and the speed of the car is increasing at a constant rate. At time 0 seconds, the speed of the car is V0 meters per second; 10 seconds later, the front bumper of the car has traveled 125 meters and the speed of the car is V10 meters per second.

In the table below, select values of V0 and V10 that are together consistent with the information provided. Make only two selections, one in each column.

My doubt is:
The question indicates that the speed of the car is increasing at a constant rate. So, is it talking about an arithmetic sequence, or a geometric sequence? According to the OE, the average speed is $$\frac{1}{2}*( V0 + V10)$$. So, it seems that it is talking about an arithmetic sequence because that's the way we use to calculate the average in an arithmetic sequence. Please confirm.
However, I remember that, when a question mentions that something is increasing at a constant rate, we must multiply the first value by a constant, we shouldn't add. Please, your help.

OA:
[Reveal] Spoiler:
V0 = 5 ; V10 = 20

t=0 speed V0
t=10 " V10

speed is raised at const rate. SO, V0, V1 ,....V10 are in AM
Let V10= V0 + (10-1)X .....X is increment for every sec

Since d=SXt
125= (V0(1sec) +V1(1sec) +...V10(1sec

So, 125= V0+ V0+1X + V0+2X +...V0+9X = 10V0 + (1+2+3+..9)X= 10V0 + (9*5)X
But, 9X=V10-V0
So, 125 = 5V0+5V10.................V0+v10=25 (5+20)
Current Student
Joined: 02 Oct 2012
Posts: 13
Location: United States
Concentration: Entrepreneurship, General Management
GMAT 1: 720 Q49 V38
GPA: 2.77
WE: General Management (Education)
Followers: 0

Kudos [?]: 6 [0], given: 12

### Show Tags

21 Sep 2013, 21:39
mbmanoj wrote:
danzig wrote:
A car is traveling on a straight stretch of roadway, and the speed of the car is increasing at a constant rate. At time 0 seconds, the speed of the car is V0 meters per second; 10 seconds later, the front bumper of the car has traveled 125 meters and the speed of the car is V10 meters per second.

In the table below, select values of V0 and V10 that are together consistent with the information provided. Make only two selections, one in each column.

My doubt is:
The question indicates that the speed of the car is increasing at a constant rate. So, is it talking about an arithmetic sequence, or a geometric sequence? According to the OE, the average speed is $$\frac{1}{2}*( V0 + V10)$$. So, it seems that it is talking about an arithmetic sequence because that's the way we use to calculate the average in an arithmetic sequence. Please confirm.
However, I remember that, when a question mentions that something is increasing at a constant rate, we must multiply the first value by a constant, we shouldn't add. Please, your help.

OA:
[Reveal] Spoiler:
V0 = 5 ; V10 = 20

t=0 speed V0
t=10 " V10

speed is raised at const rate. SO, V0, V1 ,....V10 are in AM
Let V10= V0 + (10-1)X .....X is increment for every sec

Since d=SXt
125= (V0(1sec) +V1(1sec) +...V10(1sec

So, 125= V0+ V0+1X + V0+2X +...V0+9X = 10V0 + (1+2+3+..9)X= 10V0 + (9*5)X
But, 9X=V10-V0
So, 125 = 5V0+5V10.................V0+v10=25 (5+20)

Doubt :
I agree with
"Speed is raised at const rate. SO, V0, V1 ,....V10 are in AM"
But not with
"Let V10= V0 + (10-1)X .....X is increment for every sec," which will give V1=V0 for t=1 seconds

i.e. V1=V0+X, V2=V0+2X, V3 = V0+3X, so.. V10 = V0+10X

Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 3070
Followers: 1025

Kudos [?]: 4447 [3] , given: 49

### Show Tags

23 Oct 2013, 11:56
3
KUDOS
Expert's post
danzig wrote:
My doubt is:
The question indicates that the speed of the car is increasing at a constant rate. So, is it talking about an arithmetic sequence, or a geometric sequence? According to the OE, the average speed is $$\frac{1}{2}*( V0 + V10)$$. So, it seems that it is talking about an arithmetic sequence because that's the way we use to calculate the average in an arithmetic sequence. Please confirm.
However, I remember that, when a question mentions that something is increasing at a constant rate, we must multiply the first value by a constant, we shouldn't add. Please, your help.

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 2322 times ]

The diagonal dark green line is the graph of the speed vs. time for this object. The slope of this line is the acceleration. On a speed vs. time graph, the area under the curve equals the distance traveled, so that brick-red region should have an area equal to the total distance traveled. That brick-red region is a trapezoid, and
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
_________________

Mike McGarry
Magoosh Test Prep

Intern
Joined: 11 Sep 2014
Posts: 8
Followers: 0

Kudos [?]: 2 [0], given: 5

### Show Tags

28 Sep 2014, 05:57
mikemcgarry wrote:
danzig wrote:
My doubt is:
The question indicates that the speed of the car is increasing at a constant rate. So, is it talking about an arithmetic sequence, or a geometric sequence? According to the OE, the average speed is $$\frac{1}{2}*( V0 + V10)$$. So, it seems that it is talking about an arithmetic sequence because that's the way we use to calculate the average in an arithmetic sequence. Please confirm.
However, I remember that, when a question mentions that something is increasing at a constant rate, we must multiply the first value by a constant, we shouldn't add. Please, your help.

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

The diagonal dark green line is the graph of the speed vs. time for this object. The slope of this line is the acceleration. On a speed vs. time graph, the area under the curve equals the distance traveled, so that brick-red region should have an area equal to the total distance traveled. That brick-red region is a trapezoid, and
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

Awesome explanation! Thank you!
_________________

What we think, we become

Re: IR - speed   [#permalink] 28 Sep 2014, 05:57
Similar topics Replies Last post
Similar
Topics:
MSR on IR 0 15 May 2016, 12:07
2 IR Practice 4 16 Mar 2015, 05:35
1 IR Doubt Speed Distance Time 1 07 Jul 2014, 04:35
IR Fail 0 06 May 2013, 00:50
ir query 1 03 Feb 2013, 10:16
Display posts from previous: Sort by