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

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NitinPant

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06 Apr 2025, 08:40

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Just need to use v=u+at and v^2-u^2=2as

v = final speed
u= initial speed
a = acceleration
t = time
s = distance traveled

imhimanshu

A car is traveling on a straight stretch of roadway, and the speed of the car is increasing at a constant rate with respect to time. At time 0 seconds, the speed of the car is $v_0$ meters per second; 10 seconds later, the front bumper of the car has traveled 125 meters, and the speed of the car is $v_{10}$ meters per second.

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

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consistentprep

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23 Jul 2025, 05:36

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Can be solved by class 8 physics formula: U is initial velocity and V is final velocity
Formula 1 : V= U+at ==> a = (V-U)/t

Formula 2 : V^2 -U^2 = 2 a S

Solving :
(V+U)(V-U) = 2 a S = 2 (V-U)/t * 125 (cancelling V-U both sides)
V+U = 2*125/10 = 25

Also since it is accelerating final velocity V will be greater than initial velocity U

Hence the only option is V=20 and U = 5

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14 Sep 2025, 13:08

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It is just sum of arithmetic series.

Sn=[(v0+v10)/2]*n

So, [(v0+v10)/2]*10=125, then v0+v10=25. It is only possible here when v0=5 and v10=20

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24 Feb 2026, 23:40

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This approach is incorrect, and it cost me some time to figure out why, sharing here for anyone who gets confused like me.
At the end, v0+9d has been replaced with v10 but it should actually be v9. Why --

The approach starts with this -
v0 + (v0+d) + (v0+2d)+.........+(v0+8d)+(v0+9d) = 10/2[2v0 + (10-1)d]=10/2[2v0 + 9d]

and here the terms are v0, v1, v2.....v9 -- 10 terms total -- 10 segments of distances travelled.

fameatop

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.

egmat

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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

heeeya

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?

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25 Mar 2026, 11:27

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I see a lot of detailed / complex explanation, Here is a simple breakdown

Speed - > increases constantly by some value "d"

therefore, Total Distance = t0*v0 + t1*v1....... + t10*v10 =S10

Since time increases by 1 sec, and speed is constantly increases, it forms an AP.

v1 = v0 + k
v2 = v1 + k = v0 + (2-1)k
... vn = vo + (n-1)k

total d = 125 = S10
Apply AP Sum formula

10/2 * (v0 + v10) = 125

v0 + v10 = 25 --- only possible values are 5, 20

egmat

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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 + 2d, v(3) = v0 + 3d, 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 [b]125 meters:[/b]
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

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03 Apr 2026, 08:25

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This is typical sum of series formula applied:
n/2[a+l] = 125
[a+l] = 125/(10/2) = 25
Find a,l where a+l = 25.
Answer: 5,20

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Edit: Not really sure why many are applying equations of motion here. GMAT doesn't need one to know them.

imhimanshu

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