Explanation
Lets initial velocity of car is $$v$$, final velocity of this car will be $$5\ v$$
$$K.E = \dfrac{1}{2}$$ $$ mv ^{2}$$
$$\therefore \ K.E\propto v^2$$
Given that, speed of particle is becoming 3 times of its initial value while moving. ie $$v'=3v$$
As we know $$K.E = \dfrac{1}{2}$$ $$ mv ^{2}$$ ..........(1)
After velocity tripled $$K.E' = \dfrac{1}{2}$$ $$ m(3v) ^{2}=\dfrac92mv^2$$...........(2)
From equation 1 and 2
$$K.E' = 9 \times K.E$$
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