a particle moves along the X- axis as x=u(t-2s)+a(t-2s)^2 a) the initial velocity of the particle is u. b) the acceleration of the particle is a c) the acceleration of the particle is 2a. d) at t=2s particle is at the origin the ques is multi correct type

To analyze the motion of the particle described by the equation \( x = u(t - 2s) + a(t - 2s)^2 \), we need to break down the components of the equation and understand what each term represents. Let's go through the options one by one to determine which statements are correct.

Understanding the Equation

The equation given is a quadratic function of time, where:

• x is the position of the particle along the X-axis.
• u represents the initial velocity of the particle.
• a is the acceleration of the particle.
• t is the time in seconds.

Breaking Down the Motion

The term \( (t - 2s) \) indicates that the motion is being observed starting from \( t = 2s \). This means that the particle's behavior is defined relative to this time point. Let's analyze the options:

Option a: The initial velocity of the particle is u.

When \( t = 2s \), the term \( (t - 2s) \) becomes zero. Thus, the position \( x \) at \( t = 2s \) is:

\( x = u(0) + a(0)^2 = 0 \).

However, the initial velocity \( u \) is indeed the coefficient of the linear term in the equation, which indicates that the particle starts with this velocity at \( t = 2s \). Therefore, this statement is correct.

Option b: The acceleration of the particle is a.

The acceleration can be found by taking the second derivative of the position function with respect to time. The first derivative gives us the velocity:

\( v = \frac{dx}{dt} = u + 2a(t - 2s) \).

Taking the derivative of the velocity gives:

\( a = \frac{dv}{dt} = 2a \).

This shows that the acceleration is indeed \( 2a \), not \( a \). Thus, this statement is incorrect.

Option c: The acceleration of the particle is 2a.

As derived above, the second derivative confirms that the acceleration is \( 2a \). Therefore, this statement is correct.

Option d: At t = 2s, the particle is at the origin.

As calculated earlier, when \( t = 2s \), the position \( x \) is:

\( x = u(0) + a(0)^2 = 0 \).

This means that the particle is indeed at the origin at \( t = 2s \). Therefore, this statement is also correct.

Summary of Findings

Based on the analysis, the correct options are:

• Option a: The initial velocity of the particle is u. (Correct)
• Option b: The acceleration of the particle is a. (Incorrect)
• Option c: The acceleration of the particle is 2a. (Correct)
• Option d: At t = 2s, the particle is at the origin. (Correct)

In conclusion, the correct answers are options a, c, and d. This exercise illustrates how to interpret motion equations and derive meaningful insights about a particle's behavior over time.