Question

two motorcycles are heading north. one motorcycle is 0.25 km ahead of the other motorcycle. the velocity of the motorcycles are 20.0 m/s n and 25.0 m/s n. if the faster motorcycle is behind the slower motorcycle, how long will it take for the motorcycles to meet?

Explanation:

Step1: Convert distance to SI unit

First, convert the distance of $0.25$ km to meters. Since $1$ km = $1000$ m, then $d = 0.25\times1000=250$ m.

Step2: Find relative - velocity

The relative velocity of the faster motorcycle with respect to the slower one is $v = v_2 - v_1$, where $v_2 = 25.0$ m/s and $v_1 = 20.0$ m/s. So $v=(25.0 - 20.0)$ m/s$ = 5.0$ m/s.

Step3: Calculate time

We know that $v=\frac{d}{t}$, so $t=\frac{d}{v}$. Substitute $d = 250$ m and $v = 5.0$ m/s into the formula, we get $t=\frac{250}{5.0}=50$ s.

Answer:

50 s