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if the speed of a car is increased by 65%, by what factor will its minimum braking distance be increased, assuming all else is the same? ignore the driver's reaction time.



Answer :

If the speed of the car is increased by 65%, the stopping distance of the car will be increased by 272.25%.

Let us say, the car is moving with velocity U, the stopping time of the car is T and the stopping distance of the block is d,

Now, we should use the equation of motion,

U²-V² =2aD

Where,

U is initial velocity,

V is final velocity,

D is the stopping distance,

a is the acceleration.

We know, at the final velocity will be zero,

So,

U² = 2aD

U =√(2aD)

Now, if the speed is increased by 65%, let us say the stopping distance becomes D'.

So,

U+65%U = √(2aD')

165U/100 = √(2aD')

Now, putting U = √(2aD),

165√(2aD)/100 = √(2aD')

(1.65)²D = D'

D' = 2.7225D

Now,

∆D = D' - D/D × 100

∆D = 2.7225 × 100

∆D = 272.25%

So, the stopping distance increased by 272.25%.

To know more about stopping distance, visit,

https://brainly.com/question/26711747

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