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Physics Question 28 – JEE-MAIN 2025

A particle moves along the x-axis and has its displacement x varying with time t according to the equation: x=c0(t22)+c(t2)2 where c0 and c are constants of appropriate dimensions. Then, which of the following statements is correct?

Recall that velocity is the rate of change of displacement, and acceleration is the rate of change of velocity.

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Ninja StrategyTerm-by-Term Differentiation

Recognize that the acceleration is the second derivative of displacement, and since the displacement equation is a sum of two terms, the acceleration will be the sum of the second derivatives of each term.

Step 1: Calculate the velocity of the particle.✦ Active

The displacement of the particle is given by x=c0(t22)+c(t2)2. To find the velocity v, we differentiate x with respect to time t:

v=dxdt=ddt[c0(t22)+c(t2)2] v=c0(2t)+c2(t2)1 v=2c0t+2c(t2)

The initial velocity (at t=0) is v(0)=2c0(0)+2c(02)=4c. Therefore, option 1 is incorrect.

Step 2: Calculate the acceleration of the particle.○ Expand

To find the acceleration a, we differentiate the velocity v with respect to time t:

a=dvdt=ddt[2c0t+2c(t2)] a=2c0(1)+2c(1) a=2c0+2c a=2(c0+c)
💡 Teacher's Secret Hint

Notice that the acceleration is constant, as expected for a displacement equation that is quadratic in time.

Step 3: Compare the result with the given options.○ Expand

The calculated acceleration is a=2(c0+c). Comparing this with the given options: Option 2: "the acceleration of the particle is 2c0" (Incorrect) Option 3: "the acceleration of the particle is 2c" (Incorrect) Option 4: "the acceleration of the particle is 2(c+c0)" (Correct) Thus, the correct statement is option 4.

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