KINEMATICS BYU 2

Solution 2: Duration of Rain

Let $t$ be the duration of the rain. We are given:

• Scheduled speed $v_0 = 70$ km/h.
• Speed during rain $v_1 = 60$ km/h.
• Speed after rain $v_2 = 75$ km/h.
• Distance covered after rain $s = 40$ km.

Step 1: Formulate the Time Equation
The problem states that the bus covered the remaining distance exactly in the scheduled time. This implies that the total time taken for the “Rain Part” plus the “Recovery Part” is equal to the time it would have taken to cover that specific total distance at the constant scheduled speed $v_0$.

Total Distance of interest $D = (\text{Distance in rain}) + s$.
$D = v_1 t + s$.

Step 2: Equate Actual Time and Scheduled Time
$$ T_{actual} = T_{scheduled} $$ Actual time is the sum of time spent in rain ($t$) and time spent covering the remaining distance $s$: $$ T_{actual} = t + \frac{s}{v_2} $$ Scheduled time is the total distance divided by the scheduled speed $v_0$: $$ T_{scheduled} = \frac{D}{v_0} = \frac{v_1 t + s}{v_0} $$

Step 3: Solve for $t$
Equating the expressions: $$ t + \frac{s}{v_2} = \frac{v_1 t}{v_0} + \frac{s}{v_0} $$ Group the terms containing $t$ on the left and terms with $s$ on the right: $$ t – \frac{v_1 t}{v_0} = \frac{s}{v_0} – \frac{s}{v_2} $$ $$ t \left( 1 – \frac{v_1}{v_0} \right) = s \left( \frac{1}{v_0} – \frac{1}{v_2} \right) $$ $$ t \left( \frac{v_0 – v_1}{v_0} \right) = s \left( \frac{v_2 – v_0}{v_0 v_2} \right) $$ Cancel $v_0$ from the denominator on both sides: $$ t (v_0 – v_1) = s \frac{v_2 – v_0}{v_2} $$ $$ t = \frac{s (v_2 – v_0)}{v_2 (v_0 – v_1)} $$