Question

If the speed of a vehicle increases by 22%, by what factor does its kinetic energy increase?

Answer 1

Answer:

The kinetic increases by 48.84 %

Explanation:

The expression for the kinetic energy is:-

$K.E.=\frac{1}{2}\times mv^2$

Where, m is the mass of the object

v is the velocity of the object

Let the new velocity is:- v'

v is increased by 22 %. Thus, v' = 1.22 v

So, the new kinetic energy is:-

$K.E.'=\frac{1}{2}\times mv'^2=\frac{1}{2}\times m{1.22}^2=\frac{1}{2}\times mv^21.4884=1.4884K.E.$

Thus, the kinetic increases by 48.84 %

Answer 2

Answer:

the kinetic energy of the vehicle increases 48.84% after increasing its initial speed by 22%.

Explanation:

• KE = (1/2)mv²

∴ KE1 = (1/2)mv²

⇒ KE1 = (0.5)mv²

∴ KE2 = (1/2)m(v + 0.22v)²

⇒ KE2 = (1/2)m(1.22v)²

⇒ KE2 = (1/2)m(1.4884v²)

⇒ KE2/KE1 = ((1/2)m(1.4884v²))/((1/2)mv²)

⇒ KE2/KE1 = 1.4884 = 1 + 0.4884

⇒ % increase = (0.4884)×100 = 48.84%