Toy Car \( A \) and Toy Car \( B \) are 100 m apart. Toy Car \( A \) travels at a speed of \( a \mathrm{~m} / \mathrm{s} \) while Toy Car \( B \) travels at a speed of \( b \mathrm{~m} / \mathrm{s} \). (a) If the two cars travel towards each other, they will meet in 10 seconds. Form an equation, in terms of \( a \) and \( b \), and show that it simplifies to \( a+b=10 \). (b) However, if both travels in the same direction, Toy Car \( A \) will catch up with Toy Car \( B \) in 50 seconds. Form another equation, in terms of \( a \) and \( b \), and show that it simplifies to \( a-b=2 \). [1] (c) Hence, find the respective travel speeds of Toy Car \( A \) and of Toy Car \( B \). (1]

Let's solve each part of the problem step by step. --- ### **Part (a): Meeting When Traveling Towards Each Other** **Given:** - Initial distance between Toy Car \( A \) and Toy Car \( B \) = 100 meters. - Speed of Toy Car \( A \) = \( a \) m/s. - Speed of Toy Car \( B \) = \( b \) m/s. - Time to meet = 10 seconds. **Approach:** When the two cars travel towards each other, their relative speed is the sum of their individual speeds, \( a + b \). They will meet when the combined distance they cover equals the initial separation of 100 meters. **Equation Formation:** \[ \text{Distance} = \text{Relative Speed} \times \text{Time} \] \[ 100 = (a + b) \times 10 \] **Simplification:** \[ a + b = \frac{100}{10} = 10 \] \[ \boxed{a + b = 10} \] --- ### **Part (b): Catching Up When Traveling in the Same Direction** **Given:** - Initial distance between Toy Car \( A \) and Toy Car \( B \) = 100 meters. - Speed of Toy Car \( A \) = \( a \) m/s. - Speed of Toy Car \( B \) = \( b \) m/s. - Time for Toy Car \( A \) to catch up with Toy Car \( B \) = 50 seconds. **Approach:** When both cars travel in the same direction, Toy Car \( A \) needs to cover the initial 100 meters plus the distance Toy Car \( B \) covers in the same time. The effective speed at which \( A \) catches up to \( B \) is the difference in their speeds, \( a - b \). **Equation Formation:** \[ \text{Distance to Catch Up} = \text{Relative Speed} \times \text{Time} \] \[ 100 = (a - b) \times 50 \] **Simplification:** \[ a - b = \frac{100}{50} = 2 \] \[ \boxed{a - b = 2} \] --- ### **Part (c): Finding the Respective Speeds of Toy Car \( A \) and Toy Car \( B \)** **Given Equations:** 1. \( a + b = 10 \)    [From Part (a)] 2. \( a - b = 2 \)   [From Part (b)] **Approach:** To find the values of \( a \) and \( b \), we can solve these two equations simultaneously. **Solving the Equations:** 1. **Add the two equations to eliminate \( b \):** \[ (a + b) + (a - b) = 10 + 2 \] \[ 2a = 12 \] \[ a = \frac{12}{2} = 6 \text{ m/s} \] 2. **Substitute \( a = 6 \) into the first equation to find \( b \):** \[ 6 + b = 10 \] \[ b = 10 - 6 = 4 \text{ m/s} \] **Final Speeds:** \[ \boxed{\text{Toy Car } A \text{ travels at } 6 \text{ m/s, and Toy Car } B \text{ travels at } 4 \text{ m/s.}} \]