Solution 8: Two Cars Race
Analysis:
We need to find the difference in distance covered $\Delta s = s_A(100) – s_B(100)$.
Let $v_A(t)$ be the velocity function of Car A.
Let $t_0$ be the time when Car A reaches 20 m/s.
Motion of Car B:
1. From $0$ to $t_0$: Identical to A ($v_B = v_A$).
2. From $t_0$ to $t_0+1$: Moves with zero acceleration ($v_B = 20$ m/s).
3. From $t_0+1$ onwards: Follows A’s profile with a delay of 1 second ($v_B(t) = v_A(t-1)$).
Step 1: Calculate Distance of B
We split the integral for $s_B(100)$ into three parts:
$$ s_B(100) = \int_0^{t_0} v_A dt + \int_{t_0}^{t_0+1} 20 dt + \int_{t_0+1}^{100} v_A(t-1) dt $$
Using substitution $u = t-1$ for the third integral:
$$ \int_{t_0+1}^{100} v_A(t-1) dt = \int_{t_0}^{99} v_A(u) du $$
So,
$$ s_B(100) = s_A(t_0) + (20 \times 1) + [s_A(99) – s_A(t_0)] $$
$$ s_B(100) = 20 + s_A(99) $$
Step 2: Calculate Difference $\Delta s$
$$ \Delta s = s_A(100) – s_B(100) $$
$$ \Delta s = s_A(100) – [20 + s_A(99)] $$
$$ \Delta s = [s_A(100) – s_A(99)] – 20 $$
Step 3: Evaluate Final Velocity
The term $[s_A(100) – s_A(99)]$ represents the distance covered by Car A in the 100th second.
The problem states Car A acquires a velocity of 50 m/s “a few seconds before $t=100$” and moves with this speed.
Therefore, in the interval $t \in [99, 100]$, $v_A = 50$ m/s.
$$ s_A(100) – s_A(99) = 50 \times 1 = 50 \text{ m} $$
Step 4: Final Calculation
$$ \Delta s = 50 – 20 = 30 \text{ m} $$
Answer: (a) $\Delta s = 30$ m