Given the displacement equation:
\( x = c_0 (t^2 - 2) + c (t - 2)^2 \)
We can expand this:
\( x = c_0 t^2 - 2c_0 + c (t^2 - 4t + 4) \)
\( x = c_0 t^2 - 2c_0 + c t^2 - 4ct + 4c \)
\( x = (c_0 + c) t^2 - 4ct + (4c - 2c_0) \)
To find the velocity, we differentiate the displacement equation with respect to time:
\( v = \frac{dx}{dt} = \frac{d}{dt} [(c_0 + c) t^2 - 4ct + (4c - 2c_0)] \)
\( v = 2(c_0 + c) t - 4c \)
To find the acceleration, we differentiate the velocity equation with respect to time:
\( a = \frac{dv}{dt} = \frac{d}{dt} [2(c_0 + c) t - 4c] \)
\( a = 2(c_0 + c) \)
The acceleration of the particle is \( 2(c_0 + c) \).
Thus, the correct statement is:
the acceleration of the particle is \( 2(c + c_0) \)
Final Answer: The final answer is the acceleration of the particle is \( 2(c + c_0) \)
To determine the correct statement about the particle's motion, we need to analyze the displacement equation given:
\(x = c_0 (t^2 - 2) + c (t - 2)^2\)
Let's proceed step-by-step to compute the velocity and acceleration of the particle:
\(v = \frac{dx}{dt} = \frac{d}{dt}[c_0 (t^2 - 2) + c (t - 2)^2]\)
\(= c_0 \cdot 2t + c \cdot 2(t - 2)\)
\(= 2c_0 t + 2c(t - 2)\)
\(= 2c_0 t + 2ct - 4c\)
\(= (2c_0 + 2c)t - 4c\)
\(a = \frac{dv}{dt} = \frac{d}{dt}[(2c_0 + 2c)t - 4c]\)
\(= 2c_0 + 2c\)
The acceleration of the particle is thus \(2(c_0 + c)\).
\(v_0 = (2c_0 + 2c)\cdot 0 - 4c = -4c\)
This confirms that the initial velocity is not \(4c\).
Thus, the correct statement is that the acceleration of the particle is \(2(c + c_0)\).
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)