If the distance s travelled by a body in time t is given by `s = a/t + bt^2` then the acceleration equals

Options

A. `(2a)/(t^3) + 2b`
B. `(2s)/(t^2)`
C. `2b - (2a)/(t^3)`
D. `s/(t^2)`

Answer

To find the acceleration of a body whose distance \( s \) travelled in time \( t \) is given by the equation:

\[ s = \frac{a}{t} + bt^2 \]

we will follow these steps:

### Step 1: Find the Velocity
The velocity \( v \) is defined as the derivative of the distance \( s \) with respect to time \( t \). Therefore, we need to differentiate \( s \):

\[ v = \frac{ds}{dt} \]

Differentiating \( s \):

\[ s = \frac{a}{t} + bt^2 \]

Using the quotient rule on \( \frac{a}{t} \) and the power rule on \( bt^2 \):

1. The derivative of \( \frac{a}{t} \) is \( -\frac{a}{t^2} \).
2. The derivative of \( bt^2 \) is \( 2bt \).

Combining these results, we have:

\[ v = -\frac{a}{t^2} + 2bt \]

### Step 2: Find the Acceleration
The acceleration \( a \) is defined as the derivative of the velocity \( v \) with respect to time \( t \):

\[ a = \frac{dv}{dt} \]

Now, we differentiate \( v \):

\[ v = -\frac{a}{t^2} + 2bt \]

Differentiating each term:

1. The derivative of \( -\frac{a}{t^2} \) is \( 2\frac{a}{t^3} \) (using the power rule).
2. The derivative of \( 2bt \) is simply \( 2b \).

Putting it all together, we have:

\[ a = 2\frac{a}{t^3} + 2b \]

### Final Result
Thus, the acceleration \( a \) is given by:

\[ a = \frac{2a}{t^3} + 2b \]

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