Question

Car A starts from rest at t = 0 and travels along a straight road with a constant of 2 m/s2
until it reaches a speed of 27 m/s. Afterwards it maintains this speed. Also, when t = 0,
car B located 2000 m down the road is travelling towards A at a constant speed of 20
m/s. Determine the distance travelled by car A when they pass each other. (ans: 1097.3
m).

Answer 1

Car A starts from rest at $t = 0$ and travels along a straight road with a constant of 2 m/s² until it reaches a speed of 27 m/s. Afterwards it maintains this speed. Also, when $t = 0$, car B located 2000 m down the road is travelling towards A at a constant speed of 20 m/s. Determine the distance travelled by car A when they pass each other. (ans: 1097.3 m).

Solution

$$S_A = \frac{a t_1^2}{2} + V_A t$$ $$S_B = V_B (t_1 + t)$$ $$V_A = a t_1 \gg t_1 = \frac{V_A}{a}$$ $$S = S_A + S_B$$ $$S_A = \frac{V_A^2}{2a} + V_A t; \quad S_B = V_B \left(\frac{V_A}{a} + t\right).$$ $$\begin{array}{l} S = \frac{V_A^2}{2a} + V_A t + \frac{V_B V_A}{a} + V_B t. \gg t = \frac{S - \frac{V_A^2}{2a} - \frac{V_B V_A}{a}}{V_A + V_B} = \frac{2000 - \frac{27^2}{4} - \frac{20 \times 27}{2}}{27 + 20} \\ = 32.9 \, s \\ \end{array}$$ $$S_A = \frac{a t_1^2}{2} + V_A t = \frac{2 \times \left(\frac{27}{2}\right)^2}{2} + 27 \times 32.9 = 1071 \, m$$

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