To solve the question of how fast the blue car is going $2.50 \, \text{s}$ after it starts, we need to apply the equations of motion for uniform acceleration.

Step-by-step solution:

1. Given data:

   • Acceleration of the blue car, $a_b = 4.2 \, \text{m/s}^2$.
   • Time elapsed, $t = 2.50 \, \text{s}$.
   • Initial velocity, $u_b = 0 \, \text{m/s}$ (the car starts from rest).

2. Relevant equation: The velocity of an object under uniform acceleration is given by: $v = u + at$ Substituting the values: $v_b = 0 + (4.2)(2.50)$

3. Calculation: $v_b = 10.5 \, \text{m/s}$

Thus, the speed of the blue car $2.50 \, \text{s}$ after it starts is $10.5 \, \text{m/s}$.

Would you like a deeper explanation, or assistance with other parts of the problem?
Here are 5 related questions for further exploration:

1. What is the maximum velocity achieved by the blue car during its uniform acceleration phase?
2. How far does the blue car travel during its initial 5-second acceleration?
3. What is the total time the blue car takes to decelerate to rest?
4. How would you calculate the acceleration of the yellow car if it catches up to the blue car?
5. What is the total distance traveled by the yellow car if it catches the blue car at the stop?

Tip: Always break a multi-step physics problem into smaller parts and solve systematically for each phase of motion.