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Motion word problems in miles (easy) ## Motion word problems

To calculate the velocity, displacement and time of one moving object we use the following formula:

\$\$ v = d / t \$\$

The symbol d stands for the displacement/distance, t stands for the time for which the object moved. And the symbol v stands for the velocity of the object (sometimes r as rate is used instead of v). Velocity is always given in units of speed such as km/h (kilometer per hour), m/s (meter per second) or mph (miles per hour), where the numerator always represents displacement and the denominator time. From the above formula we can derive:

\$\$ d = v × t \$\$
\$\$ t = d / v \$\$

We use the average velocity/rate, which we determine as the sum of the entire distance divided by the total time:

\$\$ v = {d_1 + d_2 +… + d_n} / {t_1 + t_2 +… + t_n} \$\$

E.g. if John traveled a distance of 5 miles in 10 minutes and then another 15 miles in 20 minutes, we can find that his velocity on the first route is:

\$\$5/10 = 0.5;mi;per;min ;→;\$\$
\$\$0.5 × 60 = 30;mph\$\$

0.5 mi per min was multiplied by 60 to get the distance per hour.

John's speed in the second route was:

\$\$15/20 = 3/4 ;mi;per;min ;→;\$\$
\$\$3/4×60=45;mph\$\$

We calculate the average velocity in miles per hour (mph) as:

\$\$ {5 + 15} / {10 + 20} = 20/30 =2/3;→\$\$
\$\$ 2/3 × 60 = 40;mph \$\$

It is important to keep track of time units, as time can be given in minutes, but velocity in mph. E.g. How far does a racing car drive at a velocity of 120 mph in 10 minutes? First of all, we have to determine that 10 min is \$1/6\$ of an hour:

\$\$ d = v × t = 120 × 1/6 = 20 ; mi \$\$

Therefore, the car will travel 20 miles in 10 minutes.

## Calculation of velocity, displacement and time for multiple objects

Examples with multiple objects are based on the aforementioned formulae, which need to be put in the equation according to the calculated problem. There are almost always several ways to calculate the result.

Example:

At the same time, Robert and Jane drove into the 4 miles long tunnel from opposite sides. Robert at a velocity of 40 mph and Jane at a velocity of 60 mph. In how many minutes will they pass each other

From the d, v, t we must find the one that will be the same for both. Because Robert and Jane entered the tunnel at the same time, their time spent in the tunnel would be the same at the time of their meeting. And the sum of their distances traveled will be equal to the length of the tunnel. These are two equations that will help us calculate the result.

\$\$ d = d_R+d_J=v_R × t + v_J × t \$\$
\$\$ 4 = 40 × t + 60 × t \$\$
\$\$ 4 = 100t \$\$
\$\$ t = 0.04 h \$\$
\$\$ t = 0.04 × 60 = 2.4 ; min \$\$

The result is 2.4 min, which is 2 min 24 s.

If time shifts enter the problem, they need to be taken into account. E.g. Anna left the house on a motorbike at 9:00, at 9:10 her brother Peter drove in the car at a velocity of 60 mph and met her in 20 minutes. How fast was Anna travelling?

Again, we can determine a variable that is the same and that is the distance d (both traveled the same distance). At the same time, we know that Anna rode the motorbike a total time of 10 minutes longer than Peter: \$t_A = 10 + 20 = 30 ;min = 0.5; h\$

\$\$ d = v × t \$\$
\$\$ d_A = d_P \$\$
\$\$ v_A × t_A = v_P × t_P \$\$

Peter drove for 20 minutes, which is \$1/3\$ of hour.

\$\$ v_A × 0.5 = 60 × 1/3 \$\$
\$\$ v_A × 0.5 = 20 \$\$
\$\$ v_A = 20 ÷ 0.5 = 40 ;mph \$\$

Anna's average velocity was 40 mph.

Motion word problems in km and m (easy) ## Motion word problems

To calculate the velocity, displacement and time of one moving object we use the following formula:

\$\$ v = d / t \$\$

The symbol d stands for the displacement/distance, t stands for the time for which the object moved. And the symbol v stands for the velocity of the object (sometimes r as rate is used instead of v). Velocity is always given in units of speed such as km/h (kilometer per hour), m/s (meter per second) or mph (miles per hour), where the numerator always represents displacement and the denominator time. From the above formula we can derive:

\$\$ d = v × t \$\$
\$\$ t = d / v \$\$

We use the average velocity/rate, which we determine as the sum of the entire distance divided by the total time:

\$\$ v = {d_1 + d_2 +… + d_n} / {t_1 + t_2 +… + t_n} \$\$

E.g. if John traveled a distance of 5 km in 10 minutes and then another 15 km in 20 minutes, we can find that his velocity on the first route is

:
\$\$5/10 = 0.5;km/min ;→;\$\$
\$\$ 0.5 × 60 = 30;km/h\$\$

0.5 km per min was multiplied by 60 to get the distance per hour.

John's speed in the second route was:

\$\$15/20 = 3/4 ;km/min ;→;\$\$
\$\$3/4×60=45;km/h\$\$

We calculate the average velocity in km/h as:

\$\${5+15}/{10+20}=20/30=2/3;km/min;→;\$\$
\$\$2/3×60=40;km/h\$\$

It is important to keep track of time units, as time can be given in minutes, but velocity in km/h. E.g. How far does a car drive on the highway at a velocity of 120 km/h in 10 minutes? First of all, we have to determine that 10 min is \$1/6\$ of an hour:

\$\$ d = v × t = 120 × 1/6 = 20 ; km \$\$

Therefore, the car will travel 20 km in 10 minutes.

## Calculation of velocity, displacement and time for multiple objects

Examples with multiple objects are based on the aforementioned formulae, which need to be put in the equation according to the calculated problem. There are almost always several ways to calculate the result.

Example:

At the same time, Robert and Jane drove into the 4 km long tunnel from opposite sides. Robert at a velocity of 40 km/h and Jane at a velocity of 60 km/h. In how many minutes will they pass each other

From the d, v, t we must find the one that will be the same for both. Because Robert and Jane entered the tunnel at the same time, their time spent in the tunnel would be the same at the time of their meeting. And the sum of their distances traveled will be equal to the length of the tunnel. These are two equations that will help us calculate the result.

\$\$ d = d_R+d_J=v_R × t + v_J × t \$\$
\$\$ 4 = 40 × t + 60 × t \$\$
\$\$ 4 = 100t \$\$
\$\$ t = 0.04 h \$\$
\$\$ t = 0.04 × 60 = 2.4 ; min \$\$

The result is 2.4 min, which is 2 min 24 s.

If time shifts enter the problem, they need to be taken into account. E.g. Anna left the house on a scooter at 9:00, at 9:10 her brother Peter drove in the car at a velocity of 60 km/h and met her in 20 minutes. How fast was Anna travelling?

Again, we can determine a variable that is the same and that is the distance d (both traveled the same distance). At the same time, we know that Anna rode the scooter a total time of 10 minutes longer than Peter: \$t_A = 10 + 20 = 30;min = 0.5;h\$

\$\$ d = v × t \$\$
\$\$ d_A = d_P \$\$
\$\$ v_A × t_A = v_P × t_P \$\$

Peter drove for 20 minutes, which is \$1/3\$ of hour.

\$\$ v_A × 0.5 = 60 × 1/3 \$\$
\$\$ v_A × 0.5 = 20 \$\$
\$\$ v_A = 20 ÷ 0.5 = 40 ;km/h \$\$

Anna's average velocity was 40 km/h.

Motion word problems in miles (difficult) ## Motion word problems

To calculate the velocity, displacement and time of one moving object we use the following formula:

\$\$ v = d / t \$\$

The symbol d stands for the displacement/distance, t stands for the time for which the object moved. And the symbol v stands for the velocity of the object (sometimes r as rate is used instead of v). Velocity is always given in units of speed such as km/h (kilometer per hour), m/s (meter per second) or mph (miles per hour), where the numerator always represents displacement and the denominator time. From the above formula we can derive:

\$\$ d = v × t \$\$
\$\$ t = d / v \$\$

We use the average velocity/rate, which we determine as the sum of the entire distance divided by the total time:

\$\$ v = {d_1 + d_2 +… + d_n} / {t_1 + t_2 +… + t_n} \$\$

E.g. if John traveled a distance of 5 miles in 10 minutes and then another 15 miles in 20 minutes, we can find that his velocity on the first route is:

\$\$5/10 = 0.5;mi;per;min ;→;\$\$
\$\$0.5 × 60 = 30;mph\$\$

0.5 mi per min was multiplied by 60 to get the distance per hour.

John's speed in the second route was:

\$\$15/20 = 3/4 ;mi;per;min ;→;\$\$
\$\$3/4×60=45;mph\$\$

We calculate the average velocity in miles per hour (mph) as:

\$\$ {5 + 15} / {10 + 20} = 20/30 =2/3;→\$\$
\$\$ 2/3 × 60 = 40;mph \$\$

It is important to keep track of time units, as time can be given in minutes, but velocity in mph. E.g. How far does a racing car drive at a velocity of 120 mph in 10 minutes? First of all, we have to determine that 10 min is \$1/6\$ of an hour:

\$\$ d = v × t = 120 × 1/6 = 20 ; mi \$\$

Therefore, the car will travel 20 miles in 10 minutes.

## Calculation of velocity, displacement and time for multiple objects

Examples with multiple objects are based on the aforementioned formulae, which need to be put in the equation according to the calculated problem. There are almost always several ways to calculate the result.

Example:

At the same time, Robert and Jane drove into the 4 miles long tunnel from opposite sides. Robert at a velocity of 40 mph and Jane at a velocity of 60 mph. In how many minutes will they pass each other

From the d, v, t we must find the one that will be the same for both. Because Robert and Jane entered the tunnel at the same time, their time spent in the tunnel would be the same at the time of their meeting. And the sum of their distances traveled will be equal to the length of the tunnel. These are two equations that will help us calculate the result.

\$\$ d = d_R+d_J=v_R × t + v_J × t \$\$
\$\$ 4 = 40 × t + 60 × t \$\$
\$\$ 4 = 100t \$\$
\$\$ t = 0.04 h \$\$
\$\$ t = 0.04 × 60 = 2.4 ; min \$\$

The result is 2.4 min, which is 2 min 24 s.

If time shifts enter the problem, they need to be taken into account. E.g. Anna left the house on a motorbike at 9:00, at 9:10 her brother Peter drove in the car at a velocity of 60 mph and met her in 20 minutes. How fast was Anna travelling?

Again, we can determine a variable that is the same and that is the distance d (both traveled the same distance). At the same time, we know that Anna rode the motorbike a total time of 10 minutes longer than Peter: \$t_A = 10 + 20 = 30 ;min = 0.5; h\$

\$\$ d = v × t \$\$
\$\$ d_A = d_P \$\$
\$\$ v_A × t_A = v_P × t_P \$\$

Peter drove for 20 minutes, which is \$1/3\$ of hour.

\$\$ v_A × 0.5 = 60 × 1/3 \$\$
\$\$v_A × 0.5 = 20 \$\$
\$\$ v_A = 20 ÷ 0.5 = 40 ;mph \$\$

Anna's average velocity was 40 mph.