The distance travelled `s` in metres by a particle in `t` second is given by `s=t^(3)+2t^(2)+t`. The speed of the particle after `1s` will be

Options

A. `8cms^(-1)`
B. `6cms^(-1)`
C. `2cms^(-1)`
D. None of these

Explanation

To find the speed of the particle after 1 second, we will follow these steps:

1. **Write down the given distance function**:
The distance travelled by the particle is given by:
\[ s(t) = t^3 + 2t^2 + t \]

2. **Differentiate the distance function to find the velocity**:
The speed (or velocity) of the particle is the derivative of the distance with respect to time. We will differentiate \(s(t)\):
\[ \frac{ds}{dt} = \frac{d}{dt}(t^3 + 2t^2 + t) \]
Using the power rule of differentiation, we get:
\[ \frac{ds}{dt} = 3t^2 + 4t + 1 \]

3. **Evaluate the velocity at \(t = 1\) second**:
Now we will substitute \(t = 1\) into the velocity function:
\[ v(1) = 3(1)^2 + 4(1) + 1 \]
Simplifying this gives:
\[ v(1) = 3 \cdot 1 + 4 \cdot 1 + 1 = 3 + 4 + 1 = 8 \]

4. **State the final answer**:
Therefore, the speed of the particle after 1 second is:
\[ 8 \text{ meters per second} \]