Question 5 &6
Question 5: A car covers 42 kms in 1 hr 40 min 48 sec with its actual speed. Find the car’s actual speed.
FPSC AD IB 2018
This is the question involving speed, distance and time. We know distance can be calculated by taking product of time and speed. But before taking product both time and speed should be taken into SI (International System of Units) units. We know SI unit for time is ‘second’ i-e if time is given in hours, we will convert it into seconds. SI unit for speed is meter per second. Relation for Distance ‘S’, time ‘t’ and speed ‘v’ is:
We are given ;
S= 42 kms
But in 1 km there are 1000 meters, so
S= 42×1000=42000 meters
Time = 1 hour 40 minutes 48 second
Now since, there are 3600 seconds in one hour and 60 seconds in one minute;
T= 3600 seconde +40×60 seconds+45 second = 3600+2400+45 = 6045 seconds
Put values in the relation
Put 0.001 in place pf ‘m’ and 0.0002 in place of ‘s’, we get;
Question 6: In 9 hours a farmer traveled 61 kms. He traveled on foot partly at the rate of 4 km/hr and partly on bicycle at the rate of 9 km/hr. The distance he walked is ?
This question seems to take distance, speed and time related solution, but here we cant use S= VxT formula directly for each individual case, because of ambiguity of partial distances and times taken.
We have to start with assumption !
Let the man walks ‘x’ kms on foot. Then because total distance is 61 kms, so, the distance traveled on bicycle will be 61-x (smaller distance to be subtracted from bigger distance).
Now we have two cases for using S= VxT
Case 1: On foot:
S1= x kms
V1 = 4 km/hr
==> x = 4 x T1
==> T1= x/4 seconds
Case2: On bicycle:
S2= 61-x kms
V2= 9 km/hr
Put values of S2, V 2and T2;
==> 61-x =9 x T2
==> T2 = 61-x/9
But total time to travel all distance of 61 km is 9 hours
i-e T1 + T2 = 9 hours
Put values of T1 & T2
So, the distance on foot is 16 kms