A point at the periphery of a disc rotating about a fixed axis moves so that its speed varies with time as `v = 2 t^2 m//s` The tangential acceleration of the point at t= 1s is

Options

A. `4m //s^2`
B. `2sqrt2m //s`
C. `sqrt2 m//s^2`
D. none of these

Explanation

To find the tangential acceleration of a point at the periphery of a disc rotating about a fixed axis, we start with the given speed equation:

1. **Given Speed Equation**:
\[ v = 2t^2 \, \text{m/s} \]

2. **Understanding Tangential Acceleration**:
Tangential acceleration (\(a_t\)) is defined as the rate of change of speed with respect to time. Mathematically, this can be expressed as:
\[ a_t = \frac{dv}{dt} \]

3. **Differentiate the Speed Equation**:
We need to differentiate the speed \(v\) with respect to time \(t\):
\[ \frac{dv}{dt} = \frac{d}{dt}(2t^2) \]
Using the power rule of differentiation, we get:
\[ \frac{dv}{dt} = 2 \cdot 2t^{2-1} = 4t \]

4. **Substituting \(t = 1\) second**:
Now, we substitute \(t = 1\) second into the expression for tangential acceleration:
\[ a_t = 4t = 4 \cdot 1 = 4 \, \text{m/s}^2 \]

5. **Conclusion**:
Therefore, the tangential acceleration of the point at \(t = 1\) second is:
\[ a_t = 4 \, \text{m/s}^2 \]