A particle is projected with a speed `10sqrt(2) ms^(-1)` and at an angle `45^(@)` with the horizontal. The rate of change of speed with respect to time at `t = 1`s is `(g = 10 ms^(-2))`

Options

A. `10/(sqrt2) ms^(-2)`
B. `10 ms^(-2)`
C. zero
D. `5 ms^(-2)`

Answer

To solve the problem, we need to analyze the motion of the particle projected at an angle of 45 degrees with an initial speed of \(10\sqrt{2} \, \text{m/s}\). We will break down the steps to find the rate of change of speed with respect to time at \(t = 1 \, \text{s}\).

### Step-by-Step Solution:

1. **Identify the Components of Initial Velocity:**
The initial speed \(u = 10\sqrt{2} \, \text{m/s}\) and the angle of projection \(\theta = 45^\circ\).
- Horizontal component of velocity, \(u_x = u \cos \theta = 10\sqrt{2} \cos 45^\circ = 10\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 10 \, \text{m/s}\).
- Vertical component of velocity, \(u_y = u \sin \theta = 10\sqrt{2} \sin 45^\circ = 10\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 10 \, \text{m/s}\).

2. **Determine the Acceleration Components:**
- The horizontal motion has no acceleration (assuming no air resistance), so \(a_x = 0 \, \text{m/s}^2\).
- The vertical motion is affected by gravity, so \(a_y = -g = -10 \, \text{m/s}^2\).

3. **Calculate the Velocity Components at \(t = 1 \, \text{s}\):**
- Horizontal velocity at \(t = 1 \, \text{s}\):
\[ v_x = u_x + a_x t = 10 + 0 \cdot 1 = 10 \, \text{m/s} \]
- Vertical velocity at \(t = 1 \, \text{s}\):
\[ v_y = u_y + a_y t = 10 - 10 \cdot 1 = 0 \, \text{m/s} \]

4. **Calculate the Speed at \(t = 1 \, \text{s}\):**
The speed \(v\) is given by:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{10^2 + 0^2} = \sqrt{100} = 10 \, \text{m/s} \]

5. **Determine the Rate of Change of Speed:**
The rate of change of speed with respect to time is given by the acceleration. Since the horizontal component has no acceleration and the vertical component is affected by gravity, we can find the resultant acceleration:
\[ a = \sqrt{a_x^2 + a_y^2} = \sqrt{0^2 + (-10)^2} = 10 \, \text{m/s}^2 \]
However, since the speed is constant in the horizontal direction and decreases in the vertical direction, the net rate of change of speed at \(t = 1 \, \text{s}\) is effectively the vertical component of acceleration:
\[ \text{Rate of change of speed} = a_y = -10 \, \text{m/s}^2 \]

### Final Answer:
The rate of change of speed with respect to time at \(t = 1 \, \text{s}\) is \(-10 \, \text{m/s}^2\).