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

Question

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.

Select values of  v_0  and  v_{10}  that are together consistent with the information provided. Make only two selections, one in each column.

KarishmaB · Accepted Answer

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)

mikemcgarry · Answer

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 58555 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

fameatop · Answer

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 =10/2

& as per question the bumper has traveled 125 m at the end of 10th second. So
10/2 = 125
 = 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.

SOURH7WK · Answer

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.

EvaJager · 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).

Vips0000 · Answer

Wow this appears to be more of a physics question

Can be solved by s =(v+u)*t /2 or combining v=u+at and v^2 =u^2+2as

mbmanoj · Answer

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)

Hadrienlbb · Answer

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!

Sajjad1994 · Answer

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.

Elite097 · Answer

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)?

KarishmaB · Answer

"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.

Elite097 · Answer

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

KarishmaB · Answer

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

Purnank · Answer

Dude Physics in GMAT.
Vt = speed at t
Vo = initial speed of car
a = acceleration 
t = time
s = distance travel.
 Vt = Vo + at  and  Vt^2 = Vo^2 + 2as 
now,
 Vt^2 - Vo^2 = 2as 
 (Vt - Vo)*(Vt + Vo) = 2as 
from  Vt = Vo + at 
 Vt + Vo = 2*\frac{s}{t} 
s=125 t=10
substittue and solve 
 Vt + Vo = 25  and we know Vt>Vo
check the values in the table that will satisfy this.
Vo = 5 and V 10 = 20.

heeeya · Answer

Thanks for the helpful adive.
But I wonder why the speed at 2nd sec is V0+d not V0 x r when it is said to increase at a 'rate'.
Doesn't it meant that V1/V0 = V2/V1?

caedmego · Answer

Not an expert on this but; it worked for me thinking of it as an average,

So the average speed is 12.5 and I need a max and a min speeds so 3 speeds in total so 12.5* 3 is 37.5 minus the average speed= 25

The min speed + the max speed must be = 25 so the answer choices from there are straightforward.

egmat · Answer

When the question says speed is increasing at a  constant rate with respect to time , it means the  amount of speed added per second is the same .

Think of it this way:
- Second 1: speed increases by  1.5  m/s
- Second 2: speed increases by  1.5  m/s
- Second 3: speed increases by  1.5  m/s
- ... and so on

This is  linear growth (arithmetic) : v = v0 + (constant × time)

What you're thinking of: 
If V1/V0 = V2/V1 (constant ratio), that would be  exponential growth  - where speed multiplies by the same factor each second.

Why exponential doesn't match "constant rate":

Let's test: If v0 =  5  and speed multiplies by  1.15  each second:
- After 1 sec:  5  ×  1.15  =  5.75  (increase of  0.75 )
- After 2 sec:  5.75  ×  1.15  =  6.61  (increase of  0.86 )
- After 3 sec:  6.61  ×  1.15  =  7.60  (increase of  0.99 )

Notice: the increase keeps getting  larger  each second ( 0.75  →  0.86  →  0.99 ). That's NOT a constant rate - the rate itself is growing!

The correct interpretation: 
"Constant rate of increase" = same  amount  added each second =  linear/arithmetic growth

This is why we use: v10 = v0 + (acceleration ×  10 )

And for distance with constant acceleration:
Distance = Average Speed × Time = (v0 + v10)/ 2  ×  10  =  125

This gives us: v0 + v10 =  25

Answer: v0 = 5, v10 = 20

egmat · Answer

Hi heeeya,

Great question! This is a very common and  important  distinction in math and physics.

The phrase 'speed is increasing at a constant rate with respect to time' means the speed increases by the SAME AMOUNT every second. This is an additive (linear) relationship, not a multiplicative (geometric) one.

Think of it this way: 
-  'Constant rate of change'  = the derivative is constant = the change per second is the same.
- So if speed increases by  d  meters per second every second: v( 1 ) = v0 + d, v( 2 ) = v0 +  2 d, v( 3 ) = v0 +  3 d, etc.

What you're describing — V1/V0 = V2/V1 — would be a constant RATIO of change, which is geometric/exponential growth.  The problem would need to say something like 'speed increases by a constant percentage each second' or 'speed doubles every second' for that interpretation.

In physics, 'increasing at a constant rate with respect to time' is simply  constant acceleration . And with constant acceleration, we can use the formula:

Distance = Average Speed × Time  
 Distance = ((v0 + v10) / 2) × 10

Plugging in the given distance of  125  meters: 
 125  = ((v0 + v10) /  2 ) ×  10 
 125  =  5  × (v0 + v10)
v0 + v10 =  25

Now we just need two choices that add to  25 . Looking at our options:  5  +  20  =  25 . That gives us v0 =  5  (Row  1  for v0) and v10 =  20  (Row  3  for v10).

Quick rule of thumb for the GMAT:  'constant rate' = additive/linear. 'Constant ratio' or 'constant percent' = multiplicative/geometric.

Answer: 1A, 3B

sohumchojar4626 · Answer

Kinematic equations

v = u + at

s = ut + 1 + (at^2)/2

From these equations, we can derive

s =  /2

In this question,

u = v0 and v = v10

From the equation s =  /2,

s = 125 and t = 10

we will have final equation

v10 + v0 = 25

v10 = 25 - v0

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

\(v_{0} \)	\(v_{10} \)
		5
		18
		20
		36
		72

GMAT Two-part Analysis (TPA) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards