A man normally takes 5 hrs to travel at a certain speed from city A to B. One day, he increases his speed by 4 km/h and finds that the journey ...from A to B takes half an hour less than the normal...
A man normally takes 5 hrs to travel at a certain speed from city A to B. One day, he increases his speed by 4 km/h and finds that the journey ...
from A to B takes half an hour less than the normal time. Find the normal speed.
Let the distance from A to B be D.
==> The time is T1 = 5 hrs.
==> Then, let the speed be S1 = D/T1 = D/ 5
==> S1 = D/ 5
==> D = 5*S1 .............(1)
After increasing the speed ( the distance remains the same.)
==> S2 = S1 + 4
==> T2 = T1 - (0.5)
==> D = S2* T2 = (S1 + 4) * (5 - 1/2)
==> D = (S1 + 4)(9/2)
==> D = (9/2)S1 + 18
But D = 5S1
==> 5S1 = (9/2)S1 + 18
==> (5-9/2) S1 = 18
==> (1/2) S1 = 18
==> S1 = 36 km/ h
Then, the normal speed is 36 km / h
Let the speed with which the man normally travels be V. It takes 5 hours for him to go from A to B.
So the distance from A to B is 5V.
When the speed is increased by 4 km/h it takes .5 hours less.
So we have 5V / (V + 4) = 5 - 0.5
=> 5V / ( V + 4) = 4.5
=> 5V = 4.5 V + 18
=> .5 V = 18
=> V = 18/.5
=> V = 36 km/h.
The normal speed is 36 km/h
Let the normal speed for the journey from A to B be v
Let the distance between A ab B be s.
Therefore s = 5v.
On a particular day speed of the man = (4+v )km.
So the tine taken for the journey = 5v/(v+4) which is les than 1/ hour from normal time for the journey = 5- 1/2 = 9/2 hrs.
5v/(v+4) = 9/2
10v = 9(v+4).
10v = 9v+36.
10v-9v = 36.
v = 36km/h.
Therefore the normal speed of the journey is 36 kms/h.