Places A and B are 80 km apart from each other.A car starts from A and other from B at the same time.If they move in the same direction,they meet in 8 hours and if they move in opposite direction they meet in 1 hour and 20 minutes.find speed ?

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Answer 1 · 2018-08-02T09:35:59

The speeds are $v_{A} = \frac{70}{2} = 35 k m h^{- 1}$ $v_A=70/2=35kmh^-1$ and $v_{B} = 25 k m h^{- 1}$ $v_B=25kmh^-1$

Let the speeds of the cars be

$= v_{A} k m h^{- 1}$ $=v_Akmh^-1$

and

$= v_{B} k m h^{- 1}$ $=v_Bkmh^-1$

Firstly, for the cars moving in the same direction

Let the distance travelled by the second car be $= d k m$ $=dkm$

Then,

$80 + d = 8 v_{A}$ $80+d=8v_A$

$d = 8 v_{B}$ $d=8v_B$

Combining these $2$ $2$ equations

$80 + 8 v_{B} = 8 v_{A}$ $80+8v_B=8v_A$

$8 (v_{A} - v_{B}) = 80$ $8(v_A-v_B)=80$

$(v_{A} - v_{B}) = 10$ $(v_A-v_B)=10$ ................................ $(1)$ $(1)$

Secondly, for the cars moving in the opposite direction

Let the distance travelled by the second car be $= x k m$ $=xkm$

Then,

$80 - x = \frac{4}{3} v_{A}$ $80-x=4/3v_A$

$x = \frac{4}{3} v_{B}$ $x=4/3v_B$

Combining these $2$ $2$ equations

$80 - \frac{4}{3} v_{B} = \frac{4}{3} v_{A}$ $80-4/3v_B=4/3v_A$

$\frac{4}{3} (v_{A} + v_{B}) = 80$ $4/3(v_A+v_B)=80$

$v_{A} + v_{B} = 60$ $v_A+v_B=60$ ......................................... $(2)$ $(2)$

Solving equations $(1)$ $(1)$ and $(2)$ $(2)$

$v_{A} = \frac{70}{2} = 35 k m h^{- 1}$ $v_A=70/2=35kmh^-1$

$v_{B} = 25 k m h^{- 1}$ $v_B=25kmh^-1$