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

Answer

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

### Step 1: Write down the position functions
The position functions for the two 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 functions to find velocities
To find the velocities of both cars, we need to differentiate their position functions with respect to time \( t \).

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

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

### Step 3: Set the velocities equal to find the time
We need to find the time \( t \) when both cars have the same velocity:
\[ v_P(t) = v_Q(t) \]
Substituting the expressions we found:
\[ a + 2bt = f - 2t \]

### Step 4: Rearrange the equation
Rearranging the equation gives:
\[ 2bt + 2t = f - a \]
Factoring out \( t \) from the left side:
\[ t(2b + 2) = f - a \]

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

### Conclusion
Thus, the time at which both cars have the same velocity is:
\[ t = \frac{f - a}{2(b + 1)} \]