Solve this graph question.

Question 2)the initial velocity of a particle moving along X axis is u(at t=0 and x=0)and its acceleration a is given by a=kx.which of the following relation is correct between its velocit and position

options:

1)v^2-u^2=2kx

2)v^2=u^2+2kx

3)v^2=u^2+kx^2

4)v^2+u^2=2kx

Correct answer form graph is 5m/s.Correct answer for above question is option 3

To tackle this problem, we need to analyze the motion of a particle moving along the X-axis, where its acceleration is dependent on its position. The given acceleration is expressed as \( a = kx \), where \( k \) is a constant. This relationship indicates that the acceleration increases with the position of the particle. Let's break down the problem step by step to derive the correct relationship between velocity and position.

Understanding the Motion

We start with the fundamental equations of motion. The acceleration \( a \) can be defined as the derivative of velocity \( v \) with respect to time \( t \):

a = \frac{dv}{dt}

However, since we have acceleration as a function of position, we can use the chain rule to relate these quantities:

a = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}

Substituting the Given Acceleration

Now, substituting \( a = kx \) into our equation gives us:

v \frac{dv}{dx} = kx

This can be rearranged to separate the variables:

v dv = kx dx

Integrating Both Sides

Next, we integrate both sides. The left side integrates with respect to \( v \) and the right side with respect to \( x \):

• Left side: \int v dv = \frac{v^2}{2} + C_1
• Right side: \int kx dx = \frac{kx^2}{2} + C_2

Combining these results, we have:

\frac{v^2}{2} = \frac{kx^2}{2} + C

Where \( C \) is a constant that can be determined using initial conditions.

Applying Initial Conditions

At \( t = 0 \) and \( x = 0 \), the initial velocity \( u \) gives us:

\frac{u^2}{2} = C

Substituting this back into our equation yields:

\frac{v^2}{2} = \frac{kx^2}{2} + \frac{u^2}{2}

v^2 = u^2 + kx^2

Identifying the Correct Option

From our derived equation, we can see that the relationship between velocity and position is:

v^2 = u^2 + kx^2

This matches with option 3 from your question. Therefore, the correct answer is indeed:

Option 3: v^2 = u^2 + kx^2

Conclusion

In summary, we derived the relationship between velocity and position for a particle under the influence of position-dependent acceleration. This approach not only demonstrates the application of calculus in physics but also reinforces the importance of understanding how different variables interact in motion. If you have any further questions or need clarification on any steps, feel free to ask!