PROBLEMS ON SPEED DISTANCE AND TIME

Problem 1 :

Joe can run twice as fast as Pete. They start at the same point and run in opposite directions for 40 minutes and the distance between them is then 16 km. How fast does Joe run?

Solution:

Pete speed be x km/h.

Joe speed = 2x km/h

Distance = 16 km

Since they are running away from each other the speed will be x + 2x = 3x km/h.

Joe's speed = 2x

= 16 km/h

Problem 2 :

A car leaves a country town at 60 km per hour and 2 hours later a second car leaves the town and catches the first car in 5 hours. Find the speed of the second car.

Solution:

Let s be the speed of the second car in km/hr

Let d be the distance in km seond car travels until it catches first car

The first car head start = 60 × 2 = 120 km

First car:

d - 120 = 60 × 5

d - 120 = 300

d = 300 + 120

d = 420 km

Second car:

d = s × 5

420 = 5s

s = 84

So, the speed of the second car is 84 km/hr.

Problem 3 :

A boy cycles from his house to a friend's house at 20 km/h and home again at 25 km/h. If his round trip takes 9/10 of an hour, how far is it to his friend's house?

Solution:

Let the distance be x km.

So, 10 km distance to friend's house.

Problem 4 :

A motor cyclist makes a trip of 500 km. If he had increased his speed by 10 km/h, he could have covered 600 km in the same time. What was his original speed?

Solution:

Let s = original speed

and s + 10 = faster speed

So, original speed is 50 km/r.

Problem 5 :

Normally I drive to work at 60 km/h. If I derive at 72 km/h I cut 8 minutes off my time for the trip. What distance do I travel?

Solution:

Let d = His travel distance

Problem 6 :

My motor boat normally travels at 24 km/h in still water. One day I travelled 36 km against a constant current in a river and it took me the same time to travel 48 km with the current. How fast was the current?

Solution:

The basic formula is d = rt  

Let x = rate of current

24 = rate in still water

and 24 + x = rate downstream

24 - x = rate upstream