Two cars `P` and `Q` start from a point at the same time in a straight line and their position are represented by `x_p(t) = at + bt^2` and `x_Q (t) = ft - t^2`. At what time do the cars have the same velocity ?

Options

A. `(a + f)/(a(1 + b))`
B. `(f - a)/(2(1 + b))`
C. `(a - f)/(1 + b)`
D. `(a + f)/(2(b - 1))`

Explanation

To find the time at which the two cars \( P \) and \( Q \) have the same velocity, we will follow these steps:

### Step 1: Write the position equations
The positions of the cars are given as:
- For car \( P \):
\[ x_P(t) = at + bt^2 \]
- For car \( Q \):
\[ x_Q(t) = ft - t^2 \]

### Step 2: Differentiate the position equations to find velocity
The velocity of each car can be found by differentiating their position equations with respect to time \( t \).

- For car \( P \):
\[ v_P(t) = \frac{dx_P}{dt} = \frac{d}{dt}(at + bt^2) = a + 2bt \]

- For car \( Q \):
\[ v_Q(t) = \frac{dx_Q}{dt} = \frac{d}{dt}(ft - t^2) = f - 2t \]

### Step 3: Set the velocities equal to find the time
We need to find the time \( t \) when the velocities of both cars are equal:
\[ v_P(t) = v_Q(t) \]
This gives us the equation:
\[ a + 2bt = f - 2t \]

### Step 4: Rearrange the equation
Rearranging the equation to isolate \( t \):
\[ 2bt + 2t = f - a \]
\[ t(2b + 2) = f - a \]

### Step 5: Solve for \( t \)
Now, we can solve for \( t \):
\[ t = \frac{f - a}{2(b + 1)} \]

This is the time at which both cars have the same velocity.

### Summary of the Solution
The time \( t \) when the velocities of cars \( P \) and \( Q \) are equal is given by:
\[ t = \frac{f - a}{2(b + 1)} \]

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