A particle moves in xy plane such that `v_(x)=50-16 t` and `y=100-4t^(2)` where `v_(x)` is in m/s and y is in m. It is also known that `x=0` when `t=0`. Determine (i) Acceleration of particle (ii) Velocity of particle when `y=0`.

Answer

To solve the problem step by step, we will break it down into two parts as outlined in the question.

### Part (i): Determine the acceleration of the particle

1. **Find the acceleration in the x-direction**:
The velocity in the x-direction is given by:
\[ v_x = 50 - 16t \]
To find the acceleration \( a_x \), we differentiate \( v_x \) with respect to time \( t \):
\[ a_x = \frac{dv_x}{dt} = \frac{d(50 - 16t)}{dt} = -16 \, \text{m/s}^2 \]

2. **Find the velocity in the y-direction**:
The position in the y-direction is given by:
\[ y = 100 - 4t^2 \]
First, we find the velocity in the y-direction \( v_y \) by differentiating \( y \) with respect to \( t \):
\[ v_y = \frac{dy}{dt} = \frac{d(100 - 4t^2)}{dt} = -8t \, \text{m/s} \]

3. **Find the acceleration in the y-direction**:
Now, we differentiate \( v_y \) to find the acceleration \( a_y \):
\[ a_y = \frac{dv_y}{dt} = \frac{d(-8t)}{dt} = -8 \, \text{m/s}^2 \]

4. **Calculate the net acceleration**:
The net acceleration \( a \) can be found using the Pythagorean theorem since \( a_x \) and \( a_y \) are perpendicular:
\[ a = \sqrt{a_x^2 + a_y^2} = \sqrt{(-16)^2 + (-8)^2} = \sqrt{256 + 64} = \sqrt{320} = 17.89 \, \text{m/s}^2 \]

### Part (ii): Determine the velocity of the particle when \( y = 0 \)

1. **Find the time when \( y = 0 \)**:
Set the equation for \( y \) to zero:
\[ 0 = 100 - 4t^2 \]
Rearranging gives:
\[ 4t^2 = 100 \implies t^2 = 25 \implies t = 5 \, \text{s} \quad (\text{only positive time is considered}) \]

2. **Calculate \( v_x \) at \( t = 5 \)**:
Substitute \( t = 5 \) into the equation for \( v_x \):
\[ v_x = 50 - 16(5) = 50 - 80 = -30 \, \text{m/s} \]

3. **Calculate \( v_y \) at \( t = 5 \)**:
Substitute \( t = 5 \) into the equation for \( v_y \):
\[ v_y = -8(5) = -40 \, \text{m/s} \]

4. **Calculate the net velocity**:
The magnitude of the net velocity \( v \) is given by:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(-30)^2 + (-40)^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, \text{m/s} \]

### Final Answers:
- (i) The acceleration of the particle is \( 17.89 \, \text{m/s}^2 \).
- (ii) The velocity of the particle when \( y = 0 \) is \( 50 \, \text{m/s} \).