guys try answering this question, tmr night will give full explanation

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Ok U actually can't use. Any formula here

But at least describe how it moves properly

And also give reasoning on why U can't use any formula here.

.

For truck A, if it decelerates from 200 km per hour to 180km, the water will go to the right side , so we cant really eliminate truck A, for other trucks many people gave reasoning on why it could be the fastest

I see a big misconception here. Not only you, many are doing it as well
If you want to use v = displacement/ time, you are assuming that the velocity is constant and there is no acceleration
But then you find the acceleration for OA, taking v = 20 m/s.
Do you guys see the contradiction here?

We should apply suvat formulas
Journey OA
s = 20
u = 0
v = ?
a = ?
t = 1

s = ut + ½at²
20 = 0 + ½a
a = 40 m/s²
Assume that the acceleration is constant
Then use v = u + at
v = 0 + 40(1) = 40 m/s

If you remember this question, I think there's something interesting to discuss.

If you recalled, I said that when the car is at A, the velocity cannot be equal to 20 m/s, assuming a constant acceleration along OA. I stopped there and didn't continue with the discussion. However, if the car continues the journey (taking my case to be true), then you'll find out the OA, BC, DE will have acceleration, making B also a correct answer.

So, what will be the real case? Turns out, you'll need to have a non-uniform acceleration along journey OA, then you can have the car travelling at 20m/s at A, and later having a constant velocity throughout the journey, making option A the only correct answer.

But still, even it's possible to have the car travelling at 20m/s at A, the working did by Harraz in his Physics Spmnetic was still wrong. He didnt understand how to use v = d/t and a = (v-u)/t correctly

I'll say there's not enough information

Assuming uniform acceleration:
Journey OA
s = ut + 0.5at²
20 = 0 + 0.5a(1)²
a = 40 m/s²
v = u + at
v = 0 + 40(1) = 40 m/s

Journey AB
s = ut + 0.5at²
20 = 40(1) + 0.5a(1)²
a = -40m/s²
v = u + at
v = 40 + (-40)(1) = 0

Which means it's accelerating, decelerating and vice versa.
If that's the case, then option B is also correct.

Unless, we are assuming a non-uniform acceleration, where we don't care how the acceleration changes, but we assume this acceleration brings the velocity at point A to be 20 m/s, then it has constant velocity all the way down the journey, making option A the only answer.

from zianhao, an unknown legend

for those who don't know about history, few years back an admin of physics spmnetic banned a guy who corrected him, and its a shame to this group, and as an spmnetic admin, I apologize from our side...admin here including me can be incorrect, so pls correct us when we are wrong.

u should know that the liquid would be responding to acceleration not speed and therefore there is no way to determine which is faster, only which is speeding up, slowing down or maintaining steady speed.

sad to see majority wrong😭😭😭

Kuizz🔥

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