Question

*Two buses were coming from two different places situated just in the opposite

direction. The average speed of one bus is 5 km/hr more than that of another on
and they had started their journey in the same time. If the distance between
places is 500 km and they meet after 4 hours, find their speed.​

Answer 1

Distance between both buses = 500km

Time taken by them to meet each other = 4hr

Let speed of bus A be v km/hr.

ATQ, speed of bus B is 5km/ hr more than that of bus A.

∴ Speed of bus B will be (v + 5) km/hr.

★ Diagram :

`\boxed{\setlength{\unitlength}{1cm}\begin{picture}(6,2)\thicklines\put(1,1){\circle*{0.2}}\put(5,1){\circle*{0.2}}\put(1,1){\vector(1,0){1}}\put(5,1){\vector(-1,0){1.5}}\put(0.2,0.9){A}\put(5.5,0.9){B}\put(1.4,1.2){v}\put(3.85,1.2){v+5}\put(2.35,0.4){\sf{500\;km}}\put(2.2,0.5){\vector(-1,0){1.3}}\put(3.55,0.5){\vector(1,0){1.5}}\end{picture}}`

★ As we know that,

• Speed = Distance/Time ... (I)

Assuming both buses as point masses,

Relative speed of object A wrt object B when the object B moves in the opposite direction of A is given by

• 
  `\boxed{\bf{v_(AB)=v_A+v_B}}`

Relative speed of bus A wrt B :

`:\implies\:\sf{v_(AB)=v_A+v_B}`

`:\implies\sf\:v_(AB)=v+(v+5)`

`:\implies\bf\:v_(AB)=2v+5`

By substituting values in (I), we get

`\leadsto\sf\:v_(AB)=(d)/(t)`

`\leadsto\sf\:2v+5=(500)/(4)`

`\leadsto\sf\:2v+5=125`

`\leadsto\sf\:2v=120`

`\leadsto\sf\:v=60\:kmph`

• Speed of bus A = 60 kmph

Speed of bus B = (v + 5)

• Speed of bus B = 65 kmph

Hope It Helps!

Answer 2

Answer: 60 km/hr and 65 km/hr

Explanation:

o-----------------------------------←500→-------------------------------------o

Bus 1: rate = r + 5 Bus 2: rate = r

time = 4 time = 4

distance = d distance = 500 - d

Use distance = rate x time

Bus 1: d = 4(r + 5) Bus 2: 500 - d = 4(r)

d = 4r + 20 -d = 4r - 500

d = 500 - 4r

Solve the System of Equations using the Substitution method:

4r + 20 = 500 - 4r

8r + 20 = 500

8r = 480

r = 60

Bus 1: rate = r + 5 Bus 2: rate = r

= 60 + 5 = 60

= 65