To solve the problem, we need to analyze the motion of the particle projected at an angle. Here are the steps to find the rate of change of speed with respect to time at \( t = 1 \, \text{s} \).
### Step 1: Identify the initial conditions
The particle is projected with an initial speed \( u = 10\sqrt{2} \, \text{m/s} \) at an angle of \( 45^\circ \) with the horizontal.
### Step 2: Resolve the initial velocity into components
The initial velocity can be resolved into horizontal and vertical components:
- Horizontal component \( u_x = u \cos(45^\circ) = 10\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 10 \, \text{m/s} \)
- Vertical component \( u_y = u \sin(45^\circ) = 10\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 10 \, \text{m/s} \)
### Step 3: Analyze the motion
The horizontal motion is uniform since there is no acceleration in the horizontal direction (ignoring air resistance). The vertical motion is subject to gravitational acceleration \( g = 10 \, \text{m/s}^2 \), acting downwards.
### Step 4: Determine the velocity at \( t = 1 \, \text{s} \)
- Horizontal velocity remains constant: \( v_x = u_x = 10 \, \text{m/s} \)
- Vertical velocity at \( t = 1 \, \text{s} \) can be calculated using the equation:
\[
v_y = u_y - g t = 10 - 10 \cdot 1 = 0 \, \text{m/s}
\]
### Step 5: Calculate the resultant velocity at \( t = 1 \, \text{s} \)
The resultant velocity \( v \) at \( t = 1 \, \text{s} \) is given by:
\[
v = \sqrt{v_x^2 + v_y^2} = \sqrt{10^2 + 0^2} = 10 \, \text{m/s}
\]
### Step 6: Determine the rate of change of speed
The speed of the particle at \( t = 1 \, \text{s} \) is \( 10 \, \text{m/s} \). Since the horizontal component of velocity is constant and the vertical component changes due to gravity, we need to find the rate of change of speed with respect to time.
However, the speed is not changing in this case because:
- The horizontal component remains constant.
- The vertical component's effect on speed does not change the overall speed at that instant.
Thus, the rate of change of speed with respect to time \( \frac{du}{dt} \) is:
\[
\frac{du}{dt} = 0 \, \text{m/s}^2
\]
### Final Answer
The rate of change of speed with respect to time at \( t = 1 \, \text{s} \) is \( 0 \, \text{m/s}^2 \).
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