Question

The ratio of energy required to accelerate a car from rest to 20m/s to the energy required needed to accelerate from 20m/s to 40m/s is

A) 1:1
B) 1:3
C) 1:2
D) 1:4

ANS:B

BUT I NEED PROCESS

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Answer 1

Change in Kinetic energy is given by ,

$\sf \bigstar\ \; \blue{\Delta K=\dfrac{1}{2}m(v_F^2-v_i^2)}$

where ,

• m denotes mass of the object
• vf denotes final velocity
• vi denotes initial velocity

Energy required to accelerate from rest to 20 m/s is ,

$\sf \Delta K_1=\dfrac{1}{2}m(20^2-0^2)$

$:\implies \sf \Delta K_1=\dfrac{1}{2}m(400)$

$:\implies \sf \Delta K_1=200m$

Energy required to accelerate from 20 m/s to 40 m/s ,

$\sf \Delta K_2=\dfrac{1}{2}m(40^2-20^2)$

$:\implies \sf \Delta K_2=\dfrac{1}{2}m(1600-400)$

$:\implies \sf \Delta K_2=\dfrac{1}{2}m(1200)$

$:\implies \sf \Delta K_2=600m$

Let's calculate ratio b/w them ,

$:\implies \sf \dfrac{\Delta K_1}{\Delta K_2}=\dfrac{200m}{600m}$

$:\implies \sf \pink{\dfrac{\Delta K_1}{\Delta K_2}=\dfrac{1}{3}}\ \; \bigstar$