A particle moves along a curve so that its coordinates at time t are x = t, `y=1/2t^(2), z=1/3t^(3)`. The acceleration at t = 1 is

Options

A. j + 2k
B. j + k
C. 2j + k
D. none of these

Answer

To find the acceleration of the particle at time \( t = 1 \), we will follow these steps:

1. **Identify the position functions**: The coordinates of the particle at time \( t \) are given as:
\[ x(t) = t, \quad y(t) = \frac{1}{2}t^2, \quad z(t) = \frac{1}{3}t^3 \]

2. **Find the velocity functions**: The velocity is the first derivative of the position with respect to time \( t \). We differentiate each coordinate function:
\[ v_x(t) = \frac{dx}{dt} = \frac{d}{dt}(t) = 1 \]
\[ v_y(t) = \frac{dy}{dt} = \frac{d}{dt}\left(\frac{1}{2}t^2\right) = t \]
\[ v_z(t) = \frac{dz}{dt} = \frac{d}{dt}\left(\frac{1}{3}t^3\right) = t^2 \]

3. **Find the acceleration functions**: The acceleration is the derivative of the velocity with respect to time \( t \). We differentiate each velocity function:
\[ a_x(t) = \frac{dv_x}{dt} = \frac{d}{dt}(1) = 0 \]
\[ a_y(t) = \frac{dv_y}{dt} = \frac{d}{dt}(t) = 1 \]
\[ a_z(t) = \frac{dv_z}{dt} = \frac{d}{dt}(t^2) = 2t \]

4. **Evaluate the acceleration at \( t = 1 \)**: Now we substitute \( t = 1 \) into the acceleration functions:
\[ a_x(1) = 0 \]
\[ a_y(1) = 1 \]
\[ a_z(1) = 2(1) = 2 \]

5. **Combine the acceleration components**: The acceleration vector at \( t = 1 \) is given by:
\[ \mathbf{a}(1) = a_x(1) \hat{i} + a_y(1) \hat{j} + a_z(1) \hat{k} = 0 \hat{i} + 1 \hat{j} + 2 \hat{k} \]
\[ \mathbf{a}(1) = \hat{j} + 2 \hat{k} \]

Thus, the acceleration of the particle at \( t = 1 \) is:
\[ \mathbf{a}(1) = \hat{j} + 2 \hat{k} \]