This article deals with the speed distance time formula and its derivation. A common set of physics problems include the determination of either speed, distance, and time of something when the other two variables are available. Furthermore, these problems are of an interesting nature since they describe the very basic situations that take place regularly for many people. Moreover, in the scenarios involving speed, distance, and time, the objects move at either the constant speeds or the average speeds.

**What is Speed Distance and Time?**

Speed, distance, and time refer to the three important variables in the field of physics. Furthermore, these three variables certainly facilitate the solving of several types of problems in physics. First of all, speed refers to the measure of how quickly an object moves from one point to another. The speed of any particular object refers to the magnitude of its velocity. It is certainly a scalar quantity.

Distance refers to the amount of space which exists between any two points. It can also be explained as the amount something has moved. Furthermore, the measurement of distance takes place in units like miles, kilometers, meters, centimeters, millimeters, yards, and inches. Moreover, the distance in which something travels has a relation with the change in position.

Time refers to the progression of events. This progression is in such a manner that it goes from the past to the present and into the future. Therefore, if there is a system which is unchanging in nature, then it is timeless. Furthermore, time is not something that one can see, touch, or taste. One can only measure its passage. Scientists believe the time to be the fourth dimension of reality. Physicists make use of time to describe event sin three-dimensional space.

**Get the huge list of Physics Formulas here**

**Speed Distance Time Formula and Derivation**

There certainly exists direct relationship between the three important variables of speed, distance, and time. The speed distance time formula demonstrates the relationship between these three variables. First of all, speed is directly related to the other two variables of distance and time. This is because, speed refers to distance divided by time and it expression is below:

Speed = \(\frac{distance}{time}\), s = \(\frac{d}{t}\)

Furthermore, one can ascertain the relationship of time with the other two variables by dividing the distance with speed. Its expression is as follows:

Time = \(\frac{distance}{speed}\), t = \(\frac{d}{s}\)

Finally, to find the distance, speed is beside time. Therefore, distance certainly is speed multiplied by time.

distance = speed Ã— time, d = st

Furthermore, here

s = speed (meters/seconds)

d = distance which is traveled (meters)

t = time (seconds)

**Solved Questions**

Q1 A dog runs from one side of a park to the other side. Furthermore, the park happens to be 80.0 meters across. The dog takes exactly 16.0 seconds to cross the park. Find out the speed of the dog?

A1 The distance which the dogs travels and the time taken are available. The dogâ€™s speed is as follows:

s = \(\frac{d}{t}\)

s = \(\frac{80}{16}\)

s = 5 m/s

Hence, the speed of the dog happens to be 5 meters per second.

Q2 A golf cart is driven at its top speed of 27.0 km/h for a period of 10.0 minutes. Find out how far did the golf cart travel?

A2 Here the first step should be to change the units of the speed and time. This would ensure that the answer would be in meters. This is certainly what the question requires.

The speed is:

s = 27 km/h

s = 27 km/h Ã— 1000m/1km Ã— 1h/60 min Ã— 1 min/60 s

s = 7.5 m/s

Now one must convert the units, the speed is certainly 7.5 m/s. So, the time which the cart traveled for was:

t = 10 min

t = 10min Ã— 60s/1min

t = 600 s

The speed of the cart and the time are certainly available. Therefore, one can find the distance by using the formula:

d = st

d = (7.5 m/s) (600 s)

d = 4500 m

Hence, the distance which the golf cart traveled is 4500 m, which is definitely equal to 4.5

## Leave a Reply