The position coordinates of a particle moving in X - Y as a function of time t are `x =2t^2+6t+25`
`y = t^2+2t+1`
The speed of the object at t = 10 s is approximately

Options

A. 31 units
B. 51 units
C. 71 units
D. 81 units

Answer

To find the speed of the particle at \( t = 10 \) seconds, we will follow these steps:

### Step 1: Differentiate the position functions
The position coordinates of the particle are given by:
\[ x(t) = 2t^2 + 6t + 25 \]
\[ y(t) = t^2 + 2t + 1 \]

We need to find the velocities in the x and y directions by differentiating these functions with respect to time \( t \).

### Step 2: Find the velocity in the x-direction
To find the velocity in the x-direction, we differentiate \( x(t) \):
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}(2t^2 + 6t + 25) = 4t + 6 \]

### Step 3: Find the velocity in the y-direction
Next, we differentiate \( y(t) \):
\[ v_y = \frac{dy}{dt} = \frac{d}{dt}(t^2 + 2t + 1) = 2t + 2 \]

### Step 4: Substitute \( t = 10 \) seconds into the velocity equations
Now we will find the velocities at \( t = 10 \) seconds.

For \( v_x \):
\[ v_x(10) = 4(10) + 6 = 40 + 6 = 46 \, \text{m/s} \]

For \( v_y \):
\[ v_y(10) = 2(10) + 2 = 20 + 2 = 22 \, \text{m/s} \]

### Step 5: Calculate the magnitude of the velocity (speed)
The speed of the particle is given by the magnitude of the velocity vector, which can be calculated using the Pythagorean theorem:
\[ \text{Speed} = v = \sqrt{v_x^2 + v_y^2} \]

Substituting the values we found:
\[ v = \sqrt{(46)^2 + (22)^2} \]
Calculating \( (46)^2 \) and \( (22)^2 \):
\[ (46)^2 = 2116 \]
\[ (22)^2 = 484 \]
Now, adding these:
\[ v = \sqrt{2116 + 484} = \sqrt{2600} \]

### Step 6: Calculate the final speed
Now we compute \( \sqrt{2600} \):
\[ \sqrt{2600} \approx 51 \, \text{m/s} \]

### Final Answer
Thus, the speed of the object at \( t = 10 \) seconds is approximately \( 51 \, \text{m/s} \).

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