Two cars A and B start off to race on a straight path with initial velocities of `8 m//"sec"`. and `5 m//"sec"`. respectively. Car A moves with the uniform acceleration of `1 m//"sec"^(2)` and car B moves with uniform acceleration of `1.1 m//sec^(2)` . If both the cars reach the winning point together, determine the length of the race track. Also determine which of the cars was ahead 10 seconds before the finish.

Answer

To solve the problem, we need to find the length of the race track (S) and determine which car was ahead 10 seconds before they finished the race. We will use the equations of motion for both cars.

### Step 1: Write the equations of motion for both cars

For car A:
- Initial velocity (u_A) = 8 m/s
- Acceleration (a_A) = 1 m/s²
- Time (t) = t seconds

Using the equation of motion:
\[ S = u_A t + \frac{1}{2} a_A t^2 \]
\[ S = 8t + \frac{1}{2} (1)t^2 \]
\[ S = 8t + 0.5t^2 \] (Equation 1)

For car B:
- Initial velocity (u_B) = 5 m/s
- Acceleration (a_B) = 1.1 m/s²

Using the equation of motion:
\[ S = u_B t + \frac{1}{2} a_B t^2 \]
\[ S = 5t + \frac{1}{2} (1.1)t^2 \]
\[ S = 5t + 0.55t^2 \] (Equation 2)

### Step 2: Set the equations equal to each other

Since both cars reach the winning point together, we can set the two equations equal to each other:
\[ 8t + 0.5t^2 = 5t + 0.55t^2 \]

### Step 3: Rearrange the equation

Rearranging gives us:
\[ 8t - 5t + 0.5t^2 - 0.55t^2 = 0 \]
\[ 3t - 0.05t^2 = 0 \]

### Step 4: Factor out t

Factoring out t:
\[ t(3 - 0.05t) = 0 \]

This gives us two solutions:
1. \( t = 0 \) (not relevant for our case)
2. \( 3 - 0.05t = 0 \)

### Step 5: Solve for t

Solving for t:
\[ 0.05t = 3 \]
\[ t = \frac{3}{0.05} = 60 \text{ seconds} \]

### Step 6: Substitute t back into either equation to find S

Using Equation 1 to find S:
\[ S = 8(60) + 0.5(60^2) \]
\[ S = 480 + 0.5(3600) \]
\[ S = 480 + 1800 \]
\[ S = 2280 \text{ meters} \]

### Step 7: Determine the position of each car 10 seconds before the finish

10 seconds before the finish, the time for both cars would be:
\[ t' = 60 - 10 = 50 \text{ seconds} \]

For car A:
\[ S_A = 8(50) + 0.5(1)(50^2) \]
\[ S_A = 400 + 0.5(2500) \]
\[ S_A = 400 + 1250 = 1650 \text{ meters} \]

For car B:
\[ S_B = 5(50) + 0.5(1.1)(50^2) \]
\[ S_B = 250 + 0.5(1.1)(2500) \]
\[ S_B = 250 + 1375 = 1625 \text{ meters} \]

### Conclusion

- The length of the race track is **2280 meters**.
- 10 seconds before the finish, car A was ahead at **1650 meters**, while car B was at **1625 meters**.