Live Chat with Former Admission Directors of INSEAD | Join Chat Room to Participate
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 Your Progress
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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
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.
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.
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.
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
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 2203 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.
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.