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

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.

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.

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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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

Responding to a pm:

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)
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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.

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).

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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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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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.