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How much work is required to accelerate a 1000 kg car from 20 m/s ...

The work done is the same as the amount of energy increase. ... The change in kinetic energy denoted by ∆KE is equal to the final KE - the ... What is the force required to stop a car of mass 100kg with two passengers of ... on a body of mass 10 kg and changes its velocity from 2 metre per second to 5 ... For the first case:

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The Bicycle Problem That Nearly Broke Mathematics - Scientific ...

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The kinetic energy formula is \frac{1}{2}mv^{2}, where m is the mass and v is the velocity in km/hr.

The mass of both the objects are the same, but the velocity for the first car say v_{1}= 90 \frac{km}{hr} = 25 \frac{m}{s}

The velocity of the second car say v_{2} = 54\frac{km}{hr} = 15 \frac{m}{s}

So, we have the ratio of K.E of the two cars as

\frac{1}{2}mv_{1}^{2} : \frac{1}{2}mv_{2}^{2}

The like terms gets cancelled and we are left with v_{1}^{2} : v_{2}^{2}

Plugging in v_{1} = 25 and v_{2}= 15 we get 25^{2} : 15^{2}

= 625 : 225

= 25 : 9

Here is how you can convert km/hr to m/s

https://byjus.com/questions/how-do-you-convert-km-h-to-m-s/

Here is how you can express a ratio in the simplest form

https://www.doodlemaths.com/dmsc-article-yr5-nmd-11-2-ratio-expressing-a-given-ratio-in-its-simplest-form/