The coordinates of a moving particle at any time t are given by x = ct and y = bt. The speed of the particle at time t is given by

Options

A. 2t(c-b)
B. `sqrt((c^(2)+b^(2)))`
C. `2tsqrt((c^(2)+b^(2)))`
D. `2tsqrt((c^(2)+b^(2)))`

Answer

To find the speed of a moving particle whose coordinates at any time \( t \) are given by \( x = ct \) and \( y = bt \), we can follow these steps:

### Step 1: Determine the distance from the origin
The distance \( s \) from the origin (0, 0) to the point (x, y) can be calculated using the Pythagorean theorem:
\[ s = \sqrt{x^2 + y^2} \]

### Step 2: Substitute the expressions for \( x \) and \( y \)
We know that \( x = ct \) and \( y = bt \). Substituting these into the distance formula gives:
\[ s = \sqrt{(ct)^2 + (bt)^2} \]

### Step 3: Simplify the expression
Now, simplify the expression inside the square root:
\[ s = \sqrt{c^2t^2 + b^2t^2} \]
Factor out \( t^2 \):
\[ s = \sqrt{(c^2 + b^2)t^2} \]

### Step 4: Further simplify
Taking the square root of \( t^2 \) gives:
\[ s = t\sqrt{c^2 + b^2} \]

### Step 5: Find the speed
The speed \( v \) of the particle is defined as the rate of change of distance with respect to time, which is given by:
\[ v = \frac{ds}{dt} \]
Differentiating \( s \) with respect to \( t \):
\[ v = \frac{d}{dt}(t\sqrt{c^2 + b^2}) = \sqrt{c^2 + b^2} \]

### Final Answer
Thus, the speed of the particle at time \( t \) is:
\[ v = \sqrt{b^2 + c^2} \]

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