Question

The kinetic energy of a car is doubled when its velocity is increased by 1 m/s. Then the initial velocity of the car is

• A. \( \left(2 + \sqrt{2}\right) \, \text{ms}^{-1} \)
• B. \( \left(1 - \sqrt{2}\right) \, \text{ms}^{-1} \)
• C. \( \left(2 - \sqrt{2}\right) \, \text{ms}^{-1} \)
• D. \( \left(1 + \sqrt{2}\right) \, \text{ms}^{-1} \)

Hint:

To solve problems where kinetic energy is doubled, use the relation \( KE = \frac{1}{2}mv^2 \) and set up an equation to find the initial velocity.

Answer 1

The Correct Option is D

The kinetic energy of an object is given by: \[ KE = \frac{1}{2}mv^2 \] Where \( m \) is the mass and \( v \) is the velocity. Let the initial velocity be \( v_0 \) and the final velocity be \( v_0 + 1 \). The kinetic energy at the initial velocity is: \[ KE_1 = \frac{1}{2}mv_0^2 \] The kinetic energy at the final velocity is: \[ KE_2 = \frac{1}{2}m(v_0 + 1)^2 \] We are given that the kinetic energy doubles, so: \[ KE_2 = 2KE_1 \] Substitute the expressions for \( KE_2 \) and \( KE_1 \): \[ \frac{1}{2}m(v_0 + 1)^2 = 2 \times \frac{1}{2}mv_0^2 \] Simplifying: \[ (v_0 + 1)^2 = 2v_0^2 \] Expanding both sides: \[ v_0^2 + 2v_0 + 1 = 2v_0^2 \] Simplifying: \[ v_0^2 - 2v_0 - 1 = 0 \] Solving this quadratic equation using the quadratic formula: \[ v_0 = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} \] Thus: \[ v_0 = 1 + \sqrt{2} \, \text{ms}^{-1} \] So the initial velocity of the car is \( \boxed{1 + \sqrt{2}} \, \text{ms}^{-1} \).