Question

A driver increases his velocity from 20 km/hr to 100 km/hr. BY what factor does he increase the kinetic energy of the car with this increase in speed? Kinetic energy is 4 times greater

Kinetic energy is 16 times greater
Kinetic energy is 25 times greater
Kinetic energy is 9 times greater
Kinetic energy is 2 times greater

Answer 1

The kinetic energy of the car increases by a factor of 25.

Step-by-step explanation:

The increase in kinetic energy of the car can be determined by comparing the initial kinetic energy to the final kinetic energy. Kinetic energy is directly proportional to the square of velocity.

In this case, the velocity is increased from 20 km/hr to 100 km/hr. Let's calculate the ratio of the final kinetic energy to the initial kinetic energy.

The initial kinetic energy is given by 1/2 * (mass of the car) * (initial velocity)^2, and the final kinetic energy is given by 1/2 * (mass of the car) * (final velocity)^2.

Let's substitute the values and calculate the ratio:

Ratio = (1/2 * (mass) * (final velocity)^2) / (1/2 * (mass) * (initial velocity)^2) = (final velocity)^2 / (initial velocity)^2.

Substituting the numbers, Ratio = (100 km/hr)^2 / (20 km/hr)^2 = 10000 / 400 = 25.

Therefore, the factor by which the kinetic energy of the car increases is 25 times greater.

Answer 2

25 times greater

Step-by-step explanation:

Let the mass of the car is m

Initial speed, u = 20 km/h = 5.56 m/s

Final speed, v = 100 km/h = 27.78 m/s

The formula for the kinetic energy is given by

`K = (1)/(2)mv^(2)`

So, initial kinetic energy

`K_(i) = (1)/(2)m(5.56)^(2)`

Ki = 15.466 m

final kinetic energy

`K_(f) = (1)/(2)m(27.78)^(2)`

Kf = 385.86 m

Increase in kinetic energy is given by

=
`\left ( (K_(f))/(K_(i)) \right )`

= 385.86 / 15.466 = 25

So, the kinetic energy is 25 times greater.