Question

car A is travelling at 25 m.s-1 when it passes point X. At exactly the same time, car B, travelling towards A at 20 m.s-1, passes point Y. X and Y are 225 m apart. calculate the distance that B travels from the time it passes point Y until it meets A.

Answer 1

Car A is travelling at 25 m.s-1 when it passes point X. At exactly the same time, car B, travelling towards A at 20 m.s-1, passes point Y. X and Y are 225 m apart. calculate the distance that B travels from the time it passes point Y until it meets A.

Solution

$$S = S _ {1} + S _ {2} = 225 \mathrm {m} = \mathrm {V} _ {\mathrm {A}} \mathrm {t} + \mathrm {V} _ {\mathrm {B}} \mathrm {t} = (\mathrm {V} _ {\mathrm {A}} + \mathrm {V} _ {\mathrm {B}}) \mathrm {t} \gg \mathrm {t} = \frac {\mathrm {S}}{(\mathrm {V} _ {\mathrm {A}} + \mathrm {V} _ {\mathrm {B}})}$$

The distance that B travels from the time it passes point Y until it meets A

$$S _ {2} = \mathrm {V} _ {\mathrm {B}} \mathrm {t} = \frac {\mathrm {S V} _ {\mathrm {B}}}{(\mathrm {V} _ {\mathrm {A}} + \mathrm {V} _ {\mathrm {B}})} = \frac {225 \times 20}{20 + 25} = 100 \mathrm {m}$$