Question

If the speed of the vehicle increases by 2m/sec. Its kinetic energy is doubled, then original speed of the vehicle is
(1)(2^1/2+1)
(2)2(2^1/2+1)
(3)2(2^1/2+1)
(4)2^1/2(2^1/2+1)

Answer 1

Answer on Question 50065, Physics, Mechanics | Kinematics | Dynamics

Question:

If the speed of the vehicle increases by $2\frac{m}{s}$, its kinetic energy is doubled, then original speed of the vehicle is

1) $\left(\sqrt{2} + 1\right)\frac{m}{s}$,

2) $2\left(\sqrt{2} - 1\right)\frac{m}{s}$,

3) $2\left(\sqrt{2} + 1\right)\frac{m}{s}$,

4) $\sqrt{2}\left(\sqrt{2} + 1\right)\frac{m}{s}$.

Solution:

By the definition, kinetic energy of the vehicle is $KE_{1} = \frac{1}{2} mv^{2}$. From the condition of the question we know, that kinetic energy of the vehicle is doubled, so $KE_{2} = 2KE_{1}$. Therefore, we can write:

$$\frac{1}{2} m v_{2}^{2} = 2 \frac{1}{2} m v_{1}^{2}.$$

Because the speed of the vehicle increases by $2\frac{m}{s}$, we can write:

$$(v_{1} + 2)^{2} = 2 v_{1}^{2}.$$

Let's take the square root of both sides of equation:

$$v_{1} + 2 = \sqrt{2} v_{1}.$$

Solving this equation for $v_{1}$ we obtain the original speed of the vehicle:

$$v_{1} = \frac{2}{\sqrt{2} - 1} \cdot \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = \frac{2(\sqrt{2} + 1)}{2 - 1} = 2(\sqrt{2} + 1) \frac{m}{s}.$$

Answer:

3) $2\left(\sqrt{2} + 1\right)\frac{m}{s}$.

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