Before you get into distance, time and speed word problems, take a few minutes to read this first and understand: How to build your credit score in USA as an international student.

Problems involving Time, Distance and Speed are solved based on one simple formula.

Distance = Speed * Time

Which implies →

Speed = Distance / Time and

Time = Distance / Speed

Let us take a look at some simple examples of distance, time and speed problems.

**Example 1.** A boy walks at a speed of 4 kmph. How much time does he take to walk a distance of 20 km?

**Solution**

Time = Distance / speed = 20/4 = 5 hours.

**Example 2.** A cyclist covers a distance of 15 miles in 2 hours. Calculate his speed.

**Solution**

Speed = Distance/time = 15/2 = 7.5 miles per hour.

**Example 3.** A car takes 4 hours to cover a distance, if it travels at a speed of 40 mph. What should be its speed to cover the same distance in 1.5 hours?

**Solution**

Distance covered = 4*40 = 160 miles

Speed required to cover the same distance in 1.5 hours = 160/1.5 = 106.66 mph

Now, take a look at the following example:

**Example 4.** If a person walks at 4 mph, he covers a certain distance. If he walks at 9 mph, he covers 7.5 miles more. How much distance did he actually cover?

Now we can see that the direct application of our usual formula Distance = Speed * Time or its variations cannot be done in this case and we need to put in extra effort to calculate the given parameters.

Let us see how this question can be solved.

**Solution**

For these kinds of questions, a table like this might make it easier to solve.

Distance | Speed | Time |

d | 4 | t |

d+7.5 | 9 | t |

Let the distance covered by that person be ‘d’.

Walking at 4 mph and covering a distance ‘d’ is done in a time of ‘d/4’

IF he walks at 9 mph, he covers 7.5 miles more than the actual distance d, which is ‘d+7.5’.

He does this in a time of (d+7.5)/9.

Since the time is same in both the cases →

d/4 = (d+7.5)/9 → 9d = 4(d+7.5) → 9d=4d+30 → d = 6.

So, he covered a distance of 6 miles in 1.5 hours.

**Example 5.** A train is going at 1/3 of its usual speed and it takes an extra 30 minutes to reach its destination. Find its usual time to cover the same distance.

**Solution**

Here, we see that the distance is same.

Let us assume that its usual speed is ‘s’ and time is ‘t’, then

Distance | Speed | Time |

d | s | t min |

d | S+1/3 | t+30 min |

s*t = (1/3)s*(t+30) → t = t/3 + 10 → t = 15.

So the actual time taken to cover the distance is 15 minutes.

*Note:* Note the time is expressed in terms of ‘minutes’. When we express distance in terms of miles or kilometers, time is expressed in terms of hours and has to be converted into appropriate units of measurement.

## Solved Questions on Trains

**Example 1.** X and Y are two stations which are 320 miles apart. A train starts at a certain time from X and travels towards Y at 70 mph. After 2 hours, another train starts from Y and travels towards X at 20 mph. At what time do they meet?

**Solution**

Let the time after which they meet be ‘t’ hours.

Then the time travelled by second train becomes ‘t-2’.

Now,

Distance covered by first train+Distance covered by second train = 320 miles

70t+20(t-2) = 320

Solving this gives t = 4.

So the two trains meet after 4 hours.

**Example 2.** A train leaves from a station and moves at a certain speed. After 2 hours, another train leaves from the same station and moves in the same direction at a speed of 60 mph. If it catches up with the first train in 4 hours, what is the speed of the first train?

**Solution**

Let the speed of the first train be ‘s’.

Distance covered by the first train in (2+4) hours = Distance covered by second train in 4 hours

Therefore, 6s = 60*4

Solving which gives s=40.

So the slower train is moving at the rate of 40 mph.

## Questions on Boats/Airplanes

For problems with boats and streams,

Speed of the boat upstream (against the current) = Speed of the boat in still water – speed of the stream

[As the stream obstructs the speed of the boat in still water, its speed has to be subtracted from the usual speed of the boat]

Speed of the boat downstream (along with the current) = Speed of the boat in still water + speed of the stream

[As the stream pushes the boat and makes it easier for the boat to reach the destination faster, speed of the stream has to be added]

Similarly, for airplanes travelling with/against the wind,

Speed of the plane with the wind = speed of the plane + speed of the wind

Speed of the plane against the wind = speed of the plane – speed of the wind

Let us look at some examples.

**Example 1.** A man travels at 3 mph in still water. If the current’s velocity is 1 mph, it takes 3 hours to row to a place and come back. How far is the place?

**Solution**

Let the distance be ‘d’ miles.

Time taken to cover the distance upstream + Time taken to cover the distance downstream = 3

Speed upstream = 3-1 = 2 mph

Speed downstream = 3+1 = 4 mph

So, our equation would be d/2 + d/4 = 3 → solving which, we get d = 4 miles.

**Example 2.** With the wind, an airplane covers a distance of 2400 kms in 4 hours and against the wind in 6 hours. What is the speed of the plane and that of the wind?

**Solution**

Let the speed of the plane be ‘a’ and that of the wind be ‘w’.

Our table looks like this:

Distance | Speed | Time | |

With the wind | 2400 | a+w | 4 |

Against the wind | 2400 | a-w | 6 |

4(a+w) = 2400 and 6(a-w) = 2400

Expressing one unknown variable in terms of the other makes it easier to solve, which means

a+w = 600 → w=600-a

Substituting the value of w in the second equation,

a-w = 400

a-(600-a) = 400 → a = 500

The speed of the plane is 500 kmph and that of the wind is 100 kmph.

## More solved examples on Speed, Distance and Time

**Example 1.** A boy travelled by train which moved at the speed of 30 mph. He then boarded a bus which moved at the speed of 40 mph and reached his destination. The entire distance covered was 100 miles and the entire duration of the journey was 3 hours. Find the distance he travelled by bus.

**Solution**

Distance | Speed | Time | |

Train | d | 30 | t |

Bus | 100-d | 40 | 3-t |

Let the time taken by the train be ‘t’. Then that of bus is ‘3-t’.

The entire distance covered was 100 miles

So, 30t + 40(3-t) = 100

Solving which gives t=2.

Substituting the value of t in 40(3-t), we get the distance travelled by bus is 40 miles.

Alternatively, we can add the time and equate it to 3 hours, which directly gives the distance.

d/30 + (100-d)/40 = 3

Solving which gives d = 60, which is the distance travelled by train. 100-60 = 40 miles is the distance travelled by bus.

**Example 2.** A plane covered a distance of 630 miles in 6 hours. For the first part of the trip, the average speed was 100 mph and for the second part of the trip, the average speed was 110 mph. what is the time it flew at each speed?

**Solution**

Our table looks like this.

Distance | Speed | Time | |

1^{st} part of journey |
d | 100 | t |

2^{nd} part of journey |
630-d | 110 | 6-t |

Assuming the distance covered in the 1^{st} part of journey to be ‘d’, the distance covered in the second half becomes ‘630-d’.

Assuming the time taken for the first part of the journey to be ‘t’, the time taken for the second half becomes ‘6-t’.

From the first equation, d=100t

The second equation is 630-d = 110(6-t).

Substituting the value of d from the first equation, we get

630-100t = 110(6-t)

Solving this gives t=3.

So the plane flew the first part of the journey in 3 hours and the second part in 3 hours.

**Example 2.** Two persons are walking towards each other on a walking path that is 20 miles long. One is walking at the rate of 3 mph and the other at 4 mph. After how much time will they meet each other?

**Solution**

Distance | Speed | Time | |

First person | d | 3 | t |

Second person | 20-d | 4 | t |

Assuming the distance travelled by the first person to be ‘d’, the distance travelled by the second person is ’20-d’.

The time is ‘t’ for both of them because when they meet, they would have walked for the same time.

Since time is same, we can equate as

d/3 = (20-d)/4

Solving this gives d=60/7 miles (8.5 miles approximately)

Then t = 20/7 hours

So the two persons meet after 2 6/7 hours.

## Practice Questions for you to solve

**Problem 1:** Click here

**Answer 1:** Click here

**Problem 2:** Click here

**Answer 2:** Click here

Meera walked to school at a speed of 3 miles per hour. Once she reached the school, she

realized that she forgot to bring her books, so rushed back home at a speed of 6 miles per

hour. She then walked back to school at a speed of 4 miles per hour. All the times, she

walked in the same route.

please explain above problem

When she walks faster the time she takes to reach her home and school is lower. There is nothing wrong with the statement. They never mentioned how long she took every time.

a man covers a distance on a toy train.if the train moved 4km/hr faster,it would take 30 min. less. if it moved 2km/hr slower, it would have taken 20 min. more .find the distance.

Let the speed be x. and time be y. A.T.Q, (x+4)(y-1/2)=d and (x-2)(y+1/3)=d. Equate these two and get the answer

Could you explain how ? you have two equations and there are 3 variables.

The 3rd equation is d=xy. Now, you have 3 equations with 3 unknowns.

The variables x and y represent the usual speed and usual time to travel distance d.

Speed comes out to be 20 km/hr and the time taken is 3 hrs. The distance traveled is 60 km.

(s + 4) (t – 1/2)= st

1…new equotion = -1/2s + 4t = 2

(s – 2) (t + 1/3)= st

2…new equotion = 1/3s – 2t = 2/3

Multiply all by 6

1… -3s + 24t = 12

2… 2s – 12t = 4

Next, use elimination

t= 3

Find s:

-3s + 24t = 12

-3s + 24(3) = 12

-3s = -60

s= 20

st or distance = 3 x 20 = 60 km/h

It’s probably the average speed that we are looking for here. Ave. Speed= total distance/ total time. Since it’s harder to look for one variable since both are absent, you can use, 3d/ d( V2V3 + V1V3 + V1V2/ V1V2V3)

2 girls meenu and priya start at the same time to ride from madurai to manamadurai, 60 km away.meenu travels 4kmph slower than priya. priya reaches manamadurai and at turns back meeting meenu 12km from manamaduai. find meenu’s speed?

Hi, when the two girls meet, they have taken equal time to travel their respective distance. So, we just need to equate their time equations

Distance travelled by Meenu = 60 -12 = 48

Distance travelled by Priya = 60 + 12 = 72

Let ‘s’ be the speed of Meenu

Time taken by Meenu => t1 = 48/s

Time taken by Priya => t2 = 72/(s+4)

t1=t2

Thus, 48/s = 72/(s+4) => 24s = 192 => s = 8Km/hr

A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 KMS away from A at the same time. On the way, however the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:

Let speed of the CAR BE x kmph.

Then, speed of the train = 3/2(x)

.’. 75/x – 75/(3/2)x= 125/(10*60) — subtracting the times travelled by two them hence trains wastage time

therefore x= 120 kmph

A cyclist completes a distance of 60 km at the same speed throughout. She travels 10 km in one hour. She stops every 20 km for one hour to have a break. What are the two variables involved in this situation?

For the answer, not variables:

60km divided by 10km/h=6 hours

60 divided by 20= 3 hours

3 hours+6 hours= 9 hours

Answer: 9 hours

Let the length of the train to prod past a point be the intrinsic distance (D) of the train and its speed be S. Its speed, S in passing the electric pole of negligible length is = D/12.

The length of the platform added to the intrinsic length of the train. So, the total distance = D + 200.

The time = 20 secs.

The Speed, S = (D + 200)/20

At constant speed, D/12 = (D + 200)/20

Cross-multiplying, 20D = 12D + 200*12

20D – 12D = 200*12; 8D = 200*12

D = 200*12/8 = 300m. 4th Aug, 2018

Can anyone solve this?

Nathan and Philip agree to meet up at the park at 5:00 pm. Nathan lives 300 m due north of the park, and Philip lives 500 m due west of the park. Philip leaves his house at 4:54 pm and walks towards the park at a pace of 1.5 m/s, but Nathan loses track of time and doesn’t leave until 4:59 pm. Trying to avoid being too late, he jogs towards the park at 2.5 m/s. At what rate is the distance between the two friends changing 30 seconds after Nathan has departed?