Physics, 31.08.2019 04:30 praptibaral70

Which requires the most amount of work by the brakes of a car? a) slowing down from 100 km/h to 70 km/h. b) slowing down from 79 km/h to a stop. c) equal amounts for either.

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This is not as simple as it looks.

What quantity are we going to compare between the two cases ?
Yes, I know ... the "amount of work". But how to find that from the
numbers given in the question ?
Is it the same as the change in speed ?
Well ? Is it ?
NO. IT's NOT.

In order to reduce the car's speed, the brakes have to absorb
the KINETIC ENERGY, and THAT changes in proportion to
the SQUARE of the speed. ( KE = 1/2 m V² )

Case 'A' :
The car initially has (1/2 m) (100²)
   = (1/2m) x 10,000 units of KE.

It slows down to    (1/2 m) x (70²)
   = (1/2m) x 4,900 units of KE.

The brakes have absorbed (10,000 - 4,900) = 5,100 units of KE.

Case 'B' :
The car initially has (1/2 m) (79²)
   = (1/2m) x    6,241 units of KE.

It slows down to a stop . . . NO kinetic energy.

The brakes have absorbed all 6,241 units of KE.

Just as we suspected when we first read the problem,
the brakes do more work in Case-B, bringing the car
to a stop from 79, than they do when slowing the car
from 100 to 70 .

But when we first read the problem and formed that
snap impression, we did it for the wrong reason.
Here, I'll demonstrate:

Change Case-B. Make it "from 70 km/h to a stop".

Here's the new change in kinetic energy for Case-B:

The car initially has (1/2 m) (70²)
   = (1/2m) x    4,900 units of KE.
It slows down to a stop . . . NO kinetic energy.
The brakes have absorbed all 4,900 units of KE.

-- To slow from 100 to 70, the brakes absorbed 5,100 units of KE.

-- Then, to slow the whole rest of the way from 70 to a stop,
the brakes absorbed only 4,900 units of KE.

-- The brakes did more work to slow the car the first 30 km/hr
than to slow it the whole remaining 70 km/hr.

That's why you can't just say that the bigger change in speed
requires the greater amount of work.

It works exactly the same in the opposite direction, too.

It takes less energy from the engine to accelerate the car
from rest to 70 km/hr than it takes to accelerate it the
next 30, to 100 km/hr !

Answer from: Quest

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

$v = 5.4 t$

after 28s,

$v(28) = 5.4(28) = 151.2$ m/s

which i would round to 150 m/s as my final answer

Answer from: Quest

He can play two games. 13/4=3.253.25*2=6.5

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