How to Calculate Average Speed

This type of question, by nature, includes quite a lot of data. Therefore, the first piece of advice for you is to stick to order and organization, and prepare all the data that appears in the question in an orderly table.

Two important things:

• Placing the data in a table is highly recommended in the exam! (On the quiz or on a draft).
• Don't forget rest stops! Stoppages should also be calculated and noted (with speed = 0 and distance = 0). This is a common data point in speed questions.

Example question:

Tatiana went shopping in honor of the last day of school! She was not satisfied with going to just one mall, so she went to several different ones. First, she drove to a mall in Madrid at a speed of about 80 km/h for two hours. After the first place, she felt tired and stopped for a short time of one hour on the side of the road. After the break, she drove at a speed of about 160 km/h to the Salamanca mall for two hours. If so, what is the average speed at which Tatiana traveled?

The formula to calculate the average speed: the entire distance Tatiana traveled, divided by the total time spent.

$160+0+320=480$

The entire distance must be divided by the total time:
$2+2+1=5$

$\frac{480}{5} =96$ This is Tatiana's average speed.

With the use of the formula:

$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{480}{5} = 96 \text{ km/h}$

Additional examples:

Manuel and Gastón decided to enjoy a summer vacation in Barcelona! They left Madrid at $13:00$ at a speed of about $75$ km/h. At $15:00$ they took a one-hour break. After that, they continued driving at a speed of about $90$ km/h and arrived in Barcelona at $19:00$ . What is the average speed at which Manuel and Gastón traveled?

And now, let's calculate the average speed at which Manuel and Gastón were driving. The formula for such a calculation is to divide the distance by the total time of all the trips they made.

$150+0+270=420 \text{ km}$
Breaking down the time: $13:00$ to $15:00 = 2$ hours,$1$ hour break, \(16:00\) to $19:00 = 3$ hours. The distance must be divided by the time

$3+1+2=6 \text{ hours}$
The calculation: $\frac{420}{6}=70$ km

Using the formula:

$\text{Average Speed} = \frac{420}{6} = 70 \text{ km/h}$

Another example:

Ramiro and Roberto decided to go to the market to buy furniture for their new home! At $10:00$ they left Pescara at $85$ km/h, and arrived in Rome at $12:00$. They walked around the market for $3$ hours and bought a new table, living room set, and buffet! On the way back home, they drove at $50$ km/h due to traffic jams, and arrived only $3$ hours later. What is the average speed at which Ramiro and Roberto were driving? 

Now, let's calculate the average speed at which Ramiro and Roberto traveled:

$170+0+150=320 \text{ km}$
The distance should be divided by the time $2+3+3=8 \text{ hours}$
The calculation: $\frac{320}{8}=40$ km

With the formula:

$\text{Average Speed} = \frac{320}{8} = 40 \text{ km/h}$

At first glance, this looks like the same term, but in practice, it is not. Average speed asks you to know what is the general-classical average of the speed at which several drivers were traveling:

Mean of Speeds (Simple Average)

This is the arithmetic mean you calculate by adding speeds and dividing by how many there are.

Example:

Three drivers are traveling:

• Ivan traveled at $70$ km/h.
• Samuel at $80$ km/h.
• Robert at $120$ km/h

The average velocity of all drivers by adding the speeds and dividing - $\frac{70 + 80 + 120}{3} = \frac{270}{3} = 90 \text{ km/h}$.

Average Speed (Total Distance ÷ Total Time)

Average speed is calculated using the actual distances traveled and time taken:

$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$

This is NOT the same as the mean of speeds! Average speed accounts for how long you traveled at each speed, which makes a significant difference in the final answer.

Key Takeaway

When a problem describes a journey with different speeds, times, and distances, you must use the average speed formula, not simply average the speeds together. This is one of the most common mistakes students make on exams.

Assignment

The truck driven by Javier completes its route in two parts.

In the first part, its speed is $82$ km/h and it travels for $4$ hours.

After this part, Javier takes a break at a gas station for $20$ minutes.

In the second part, Javier travels at a speed of $70$ km/h for $3$ hours.

What is his average speed?

Solution:

The average speed is equal to the total distance divided by the total time

$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$

$part~2+part~1=The~total~route$

We calculate part $1$

Speed of part $1$ multiplied by time of part $1$ is equal to

$82\cdot4=328$

We calculate part $2$

Speed of part $2$ multiplied by time of part $2$ is equal to

$70\cdot3=210$

Total time = Time of part $1+$ break time $+$ time of part $2$

We calculate the total time

$4+\frac{1}{3}+3=7\frac{1}{3}$

We calculate the total distance traveled

$328+210=538$

The average speed is

$\frac{538}{7\frac{1}{3}}=73.36$

Answer

$73.36$

In a relay race, three runners run one after another on a track that is $450$ meters long.

The first finished in $1.5$ minutes

The second finished in $1.35$ minutes

The third finished in $1.42$ minutes

What is the average speed of the entire team?

Solution

$X_{total}=3\cdot450=1350m$

$t_{tot}=1.5+1.35+1.42=4.27\min$

$\overline{V}=\frac{1350}{4.27}=316\frac{m}{\min}=316\frac{m}{60\sec}=5.3\frac{m}{\sec}$

Answer

$5.3$ meters per second

Gastón follows the path in the figure

$A\xrightarrow{}B\xrightarrow{}C$

The path forms a right triangle.

The average speed is $2.1$ km/h

What is the speed between $C$ and $A$?

Solution

Right triangle $ABC$

Pythagorean theorem

$AB^2+BC^2=AC^2$

$5^2+4^2=AC^2$

$25+16=AC^2$

We extract the square root

$AC=\sqrt{25+16}$

$AC=\sqrt{41}$

$X_{tot}=AB+BC+CA=$

$5+4+\sqrt{41}=9+\sqrt{41}$

$t_{tot}=t_{AB}+t_{BC}+t_{CA}=$

$2+\frac{X_{BC}}{V_{BC}}+\frac{X_{AC}}{V_{AC}}=$

$2+\frac{4}{3}+\frac{\sqrt{41}}{V_{AC}}=$

$3\frac{1}{3}+\sqrt{41}$

$3\frac{1}{3}+\frac{\sqrt{41}}{V_{AC}}$

Replace in the formula:

$2.1=\frac{9+\sqrt{41}}{3\frac{1}{3}+V_{AC}}$

$3\frac{1}{3}+\frac{\sqrt{41}}{V_{AC}}=\frac{9+\sqrt{41}}{2.1}=7.335$

We subtract $3\frac{1}{3}$

$\frac{\sqrt{41}}{V_{AC}}=4.001$

We multiply by: $V_{AC},4.001$

$V_{AC}=\frac{\sqrt{41}}{4.001}=1.6\frac{km}{hr}$

Answer

$1.6$ km/h

Gerardo returns from school to his home.

On the way home, Gerardo passed by an ice cream shop.

The time it took him to go to the shop was $17$ minutes and he covered a distance of $1700$ meters.

The time it took him to get home from the shop is $20$ minutes and he covered a distance of $3000$ meters.

The average speed was $1.567$ meters per second.

How much time did he spend at the ice cream shop?

Solution

(distance in meters, times are in minutes, so units must be converted)

$\overline{V}=1.567\frac{m}{\sec}=1.567\frac{m}{\frac{1}{60}\min}=94\frac{m}{\min}$

$94=\frac{4700}{37+t}$

(t=shops)

$\left(37+t\right)94=4700$

$37\cdot94+94\cdot t=4700$

$3478+94\cdot t=4700$

We subtract $3478$

$94\cdot t=1222$

We divide by $94$

$t=\frac{1222}{94}=13\min$

Answer

$13\min$

Sergio follows a circular path with a diameter of $750$ meters, $7$ times.

The first two times the total time for the journey is $10$ minutes and a half

In the next three laps his speed is $9$ km/h

In the last two laps, the complete lap takes $12$ minutes

What is the average speed?

Solution

$7\cdot2\pi\cdot r=7\cdot2\pi\cdot\frac{diameter}{2}$

$7\cdot2\pi\cdot\frac{750}{2}=$

We simplify by: $2$

$7\cdot3.14\cdot750=16485m=16.485\operatorname{km}$

$t_{tot}=t_1+t_2+t_3+t_4+t_5+t_6+t_7$

$t_1+t_2=10.5\min=\frac{10.5}{60}=0.175hr$

$t_6+t_7=12\min=\frac{12}{60}=\frac{1}{5}=0.2hr$

$t_3=t_4=t_5=$

$\frac{\frac{2\pi\cdot r}{1000}}{9}=$

$\frac{\frac{2\cdot3.14\cdot\frac{750}{2}}{1000}}{9}$

We simplify by: $2$

$2\cdot0.175+3\cdot0.262+2\cdot0.2=1.536hr$

Answer

$10.732\frac{\operatorname{km}}{h}$

A jaguar begins to stalk a deer at $6$ in the morning, after $X$ minutes it starts to run after her at a speed of $70$ km/h for $8$ minutes.

The deer begins to accelerate and so does the jaguar for another $4$ minutes of the chase until he catches up with her.

The average speed of the jaguar from the start of the stalk to the capture is $80$ km/h.

Express using $X$ his speed in the last $4$ minutes.

Solution

$X$ plus $8$ plus $4$ minutes =

$X$ plus $12$ divided by $60$ minutes =

Replace in the formula:

$80=\frac{9\frac{1}{3}+\frac{V_1}{15}}{\frac{x+12}{60}}$

Multiplied by: $\frac{x+12}{60}$

$\frac{80}{60}\left(x+12\right)=9\frac{1}{3}+\frac{V_2}{15}$

$\frac{4}{3}x+16=9\frac{1}{3}+\frac{V_2}{15}$

Subtract $-9\frac{1}{3}$

$\frac{4}{3}x+6\frac{2}{3}=\frac{V_2}{15}$

Multiplied by $15$

$V_2=20x+100$

Answer

$100+20x$ km/h

What is average speed?

Average speed is the total distance traveled divided by the total time taken for the entire journey. It tells us how fast an object traveled overall, accounting for all speed changes, stops, and delays.

Formula:

$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$

Example:

Julian travels from one city to another in two stages. In the first stage, he travels at a speed of $110\frac{\operatorname{km}}{h}$ for 2 hours. Then he stops to eat for an hour, and in the second stage, he travels at a speed of $80\frac{\operatorname{km}}{h}$ for $3$ hours. Calculate the average speed Julian had on the trip.

Solution:

In the first stage, he travels at a speed of $110\frac{\operatorname{km}}{h}$ for two hours, so the distance covered is:

$2h\times110\frac{\operatorname{km}}{h}=220\operatorname{km}$

In the second stage, he travels at a speed of $80\frac{\operatorname{km}}{h}$ for $3$ hours. So:

$3h\times80\frac{\operatorname{km}}{h}=240\operatorname{km}$

With this, the total distance covered is:

$220\operatorname{km}+240\operatorname{km}=460\operatorname{km}$

Now let's add up the travel time:

$t_1+t_{meal}+t_3=2h+1h+3h=6h$

Now we calculate the average speed

$\frac{460\operatorname{km}}{6h}=76.7\frac{\operatorname{km}}{h}$

Result

$76.7\frac{\operatorname{km}}{h}$

How is average speed written in physics?

The average speed or mean velocity is calculated as the sum of all displacements divided by the sum of all times taken on a journey, mathematically we can express this statement as follows:

$V_m=\frac{\sum_{i\mathop{=}1}^n displacements}{\sum_{i\mathop{=}1}^n times}$

Where the numerator represents the sum of displacements and the denominator the sum of all times.

How to calculate average speed from a table?

To answer this question, let's look at the following example:

Diana studies the behavior of a particle moving in a straight line, observing that it travels at a speed of $40\frac{\operatorname{km}}{h}$ for one hour. Then it accelerates to a speed of $70\frac{\operatorname{km}}{h}$ for 3 hours and finally travels at a speed of $110\frac{\operatorname{km}}{h}$ for 5 hours. What is the particle's average speed?

Let's record these speeds and times in the following table:

So we can calculate the average speed with the table:

Total Displacement

$40\operatorname{km}+210\operatorname{km}+550\operatorname{km}=800\operatorname{km}$

Total Time

$t_1+t_2+t_3=1h+3h+5h=9h$

Therefore, the average speed is as follows:

$V_m=\frac{800\operatorname{km}}{9h}=88.9\frac{\operatorname{km}}{h}$

Result

$88.9\frac{\operatorname{km}}{h}$

What is instantaneous speed?

Instantaneous speed is the speed of an object at a specific time, this time interval is very small, meaning the time to perform this movement is extremely short (in a brief instant).

What is the difference between instantaneous speed and average speed?

As already mentioned, instantaneous speed occurs in a brief instant, in a very small amount of time, while average speed is the average of speeds that an object has over some time intervals (it is the quotient of the sum of the displacements over the sum of all the times of the movement), this interval can be much larger compared to instantaneous speed.