A car is traveling on a straight stretch of roadway, and the speed of : Two-part Analysis

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Re: A car is traveling on a straight stretch of roadway, and the speed of [#permalink]

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 traveling on a straight stretch of roadway, and the speed of [#permalink]

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 traveling on a straight stretch of roadway, and the speed of [#permalink]

17 Sep 2012, 09:08

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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 traveling on a straight stretch of roadway, and the speed of [#permalink]

17 Sep 2012, 09:24

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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 traveling on a straight stretch of roadway, and the speed of [#permalink]

31 Jul 2013, 09:52

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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:

Show 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)

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Re: A car is traveling on a straight stretch of roadway, and the speed of [#permalink]

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:

Show 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

Please comment.

S

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Re: A car is traveling on a straight stretch of roadway, and the speed of [#permalink]

24 Nov 2017, 09:59

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mikemcgarry,

Would you mind briefly commenting my following logic?

We are told that the speed of the car is increasing at a constant rate with respect to time, that time is 10 seconds and that the car has traveled 125 meters.

Therefore, I conclude that my Speed at V0 will be lower than 12.5 meters (125/10 = 12.5 meters / second). Then there is only one choice for V0 in the table. V0 = 5.

Then I know I need the rate to constantly increase until V10 and need to get to 125 meters.
(V10 + V0)/2 = 12.5 meters/s
(V10 + 5)/2 = 12.5
V10 + 5 = 12.5*2
V10 = 25 - 5 = 20

Thanks!

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

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Re: A car is traveling on a straight stretch of roadway, and the speed of [#permalink]

27 Nov 2017, 11:13

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Hadrienlbb wrote:

mikemcgarry,

Would you mind briefly commenting my following logic?

We are told that the speed of the car is increasing at a constant rate with respect to time, that time is 10 seconds and that the car has traveled 125 meters.

Therefore, I conclude that my Speed at V0 will be lower than 12.5 meters (125/10 = 12.5 meters / second). Then there is only one choice for V0 in the table. V0 = 5.

Then I know I need the rate to constantly increase until V10 and need to get to 125 meters.
(V10 + V0)/2 = 12.5 meters/s
(V10 + 5)/2 = 12.5
V10 + 5 = 12.5*2
V10 = 25 - 5 = 20

Thanks!

Dear Hadrienlbb,

My friend, I would say that yours is a brilliant and wonderfully elegant solution! Kudos!

Mike

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Re: A car is traveling on a straight stretch of roadway, and the speed of [#permalink]

25 Dec 2020, 20:42

There is a simple formula to calculate the distance traveled by object with consistently increasing speed:
S=V0t+at²/2
where S is the distance, a is the rate of speed increase, t is the time

So we get 125 = 10*V0 + a*(10)^2/2 = 10*V0 + 50a
--> 25=2V0 + 10a
And we know that V10 = V0+10a (speed after 10 seconds)
So S = V0 + V0 + 10a = V0+V10.

Only 5 and 20 will satisfy this equation.

L

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Re: A car is traveling on a straight stretch of roadway, and the speed of [#permalink]

25 Aug 2021, 08:40

Expert Reply

Official Explanation

Since the speed of the car increases at a constant rate, the car's average speed, in meters per second, over the 10-second period is equal to \(\frac{1}{2} (v_{0} + v_{10})\). Thus, the distance the car traveled, in meters, over that period is equal to \(\frac{1}{2} (v_{0} + v_{10})(10)\), or \(5(v_{0} + v_{10})\). But this value is given to be 125, so it must be true that \(v_{0} + v_{10} = \frac{125}{5}\), or 25. Because speed must be positive, both \(v_{0}\) and \(v_{10}\) must be less than 25, ruling out 36 and 72 as possible values.
Among the available alternatives—5, 18, and 20—the only pair that sum to 25 are 5 and 20. Since the speed is increasing, it must also be true that \(v_{0} ≤ v_{10}.\) Therefore \(v_{0} = 5.\)

The correct answer is 5.

Following the analysis above, the value of \(v_{10}\) must correspond to \(v_{0} = 5\), and it is given by \(v_{10} = 20.\)

The correct answer is 20.

D

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Re: A car is traveling on a straight stretch of roadway, and the speed of [#permalink]

19 Feb 2024, 05:10

KarishmaB @anishpassai pls explain this.

How do we know it is not in GP? Quest says it is increasing at 'constt growth rate'.

And since everyone has taken it in Ap, then does it mean there are a total of 11 terms from 0 secs to 10secs: v0,.......,v10. Thus shouldnt sum of ap be (11/2)(2v0+ 10d) and not (10/2)(2v0+ 9d)?

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Re: A car is traveling on a straight stretch of roadway, and the speed of [#permalink]

19 Feb 2024, 06:49

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Elite097 wrote:

KarishmaB @anishpassai pls explain this.

How do we know it is not in GP? Quest says it is increasing at 'constt growth rate'.

And since everyone has taken it in Ap, then does it mean there are a total of 11 terms from 0 secs to 10secs: v0,.......,v10. Thus shouldnt sum of ap be (11/2)(2v0+ 10d) and not (10/2)(2v0+ 9d)?

"the speed of the car is increasing at a constant rate" means the speed is increasing by a fixed amount. This 'fixed amount' is the common diff and it has to be AP.
When it is a GP, the speed increases exponentially, not at a constant rate.

D

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Re: A car is traveling on a straight stretch of roadway, and the speed of [#permalink]

10 Mar 2024, 05:45

KarishmaB yes but constant rate refers to constant percentage hence it could still increase exponentially and other questions also use growth rate to refer to GP so not clear

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Re: A car is traveling on a straight stretch of roadway, and the speed of [#permalink]

11 Mar 2024, 03:46

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Elite097 wrote:

KarishmaB yes but constant rate refers to constant percentage hence it could still increase exponentially and other questions also use growth rate to refer to GP so not clear

No. If speed is increasing at a variable rate, how does "constant rate" make sense? It means acceleration is constant, say 2m/s^2.

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Re: A car is traveling on a straight stretch of roadway, and the speed of [#permalink]

11 Mar 2024, 03:46

GMAT Two-part Analysis Questions

A car is traveling on a straight stretch of roadway, and the speed of