Question

A car is traveling at 15 m/sm/s . Part A How fast would the car need to go to double its kinetic energy

Answer 1

Answer:

21.21 m/s

Explanation:

Let KE₁ represent the initial kinetic energy.

Let v₁ represent the initial velocity.

Let KE₂ represent the final kinetic energy.

Let v₂ represent the final velocity.

Next, the data obtained from the question:

Initial velocity (v₁) = 15 m/s

Initial kinetic Energy (KE₁) = E

Final final energy (KE₂) = double the initial kinetic energy = 2E

Final velocity (v₂) =?

Thus, the velocity (v₂) with which the car we travel in order to double it's kinetic energy can be obtained as follow:

KE = ½mv²

NOTE: Mass (m) = constant (since we are considering the same car)

KE₁/v₁² = KE₂/v₂²

E /15² = 2E/v₂²

E/225 = 2E/v₂²

Cross multiply

E × v₂² = 225 × 2E

E × v₂² = 450E

Divide both side by E

v₂² = 450E /E

v₂² = 450

Take the square root of both side.

v₂ = √450

v₂ = 21.21 m/s

Therefore, the car will travel at 21.21 m/s in order to double it's kinetic energy.

Answer 2

For the car to be able to double its kinetic energy, it would need to travel at a speed of approximately 21.21m/s.

Given the data in the question;

Speed of the car; $v_1 = 15m/s$

Speed for the car to double its kinetic energy; $v_2 = \ ?$

Using the expression for kinetic energy:

$K_E = \frac{1}{2}mv^2$

Where m is the mass and v is the velocity

Now, Initial Kinetic Energy will be;

$K_E_1 = \frac{1}{2}mv_1^2 \\\\K_E_1 = \frac{1}{2}*m*(15)^2\\\\K_E_1 = \frac{1}{2}*m*225\\\\K_E_1 = 112.5*m$

For the kinetic energy to become double

$K_E_2 = 2 * K_E_1\\\\\frac{1}{2}mv_2^2 = 2 * ( 112.5 * m)\\\\\frac{1}{2}mv_2^2 = 2m( 112.5)\\\\v_2^2 = 4( 112.5)\\\\v_2^2 = 450\\\\v_2 = \sqrt{450}\\\\v_2 = 21.21m/s$

Therefore, for the car to be able to double its kinetic energy, it would need to travel at a speed of approximately 21.21m/s.

Learn more: https://brainly.com/question/999862