3 For tgeqslant 1 A particle moves along a line. the distance of the particle from 0 at time t seconds is x metres, where x=(9)/(2t)+t Find the acceleration of the particle when t=3

To find the acceleration of the particle when \( t = 3 \), we need to follow these steps:<br /><br />1. **Find the velocity function** by differentiating the position function \( x(t) \) with respect to \( t \).<br />2. **Find the acceleration function** by differentiating the velocity function with respect to \( t \).<br />3. **Evaluate the acceleration function** at \( t = 3 \).<br /><br />Given the position function:<br />\[ x(t) = \frac{9}{2t} + t \]<br /><br />### Step 1: Find the Velocity Function<br /><br />The velocity \( v(t) \) is the first derivative of the position \( x(t) \) with respect to \( t \):<br />\[ v(t) = \frac{dx}{dt} \]<br /><br />Differentiate \( x(t) \):<br />\[ x(t) = \frac{9}{2t} + t \]<br />\[ \frac{dx}{dt} = \frac{d}{dt} \left( \frac{9}{2t} + t \right) \]<br /><br />Using the power rule and the constant multiple rule:<br />\[ \frac{d}{dt} \left( \frac{9}{2t} \right) = \frac{9}{2} \cdot \frac{d}{dt} \left( t^{-1} \right) = \frac{9}{2} \cdot (-t^{-2}) = -\frac{9}{2t^2} \]<br />\[ \frac{d}{dt} (t) = 1 \]<br /><br />So,<br />\[ v(t) = -\frac{9}{2t^2} + 1 \]<br /><br />### Step 2: Find the Acceleration Function<br /><br />The acceleration \( a(t) \) is the first derivative of the velocity \( v(t) \) with respect to \( t \):<br />\[ a(t) = \frac{dv}{dt} \]<br /><br />Differentiate \( v(t) \):<br />\[ v(t) = -\frac{9}{2t^2} + 1 \]<br />\[ \frac{dv}{dt} = \frac{d}{dt} \left( -\frac{9}{2t^2} + 1 \right) \]<br /><br />Using the power rule:<br />\[ \frac{d}{dt} \left( -\frac{9}{2t^2} \right) = -\frac{9}{2} \cdot \frac{d}{dt} \left( t^{-2} \right) = -\frac{9}{2} \cdot (-2t^{-3}) = \frac{9}{t^3} \]<br />\[ \frac{d}{dt} (1) = 0 \]<br /><br />So,<br />\[ a(t) = \frac{9}{t^3} \]<br /><br />### Step 3: Evaluate the Acceleration Function at \( t = 3 \)<br /><br />Substitute \( t = 3 \) into the acceleration function:<br />\[ a(3) = \frac{9}{3^3} = \frac{9}{27} = \frac{1}{3} \]<br /><br />Therefore, the acceleration of the particle when \( t = 3 \) is:<br />\[ \boxed{\frac{1}{3}} \]