3 Ways To Calculate Velocity

As a quick example, let’s say a train accelerates at a constant rate from 30 m/s to 80 m/s. The average velocity of the train during this time is 30+802=55m/s{\displaystyle {\frac {30+80}{2}}=55m/s}.

The formula for this problem is vav=xf−xitf−ti{\displaystyle v_{av}={\frac {x_{f}-x_{i}}{t_{f}-t_{i}}}}, or “final position - initial position divided by final time - initial time. " You can also write this as vav{\displaystyle v_{av}} = Δx / Δt, or “change in position over change in time. "

Example 1: A car traveling due east starts at position x = 5 meters. After 8 seconds, the car is at position x = 41 meters. What was the car’s displacement? The car was displaced by (41m - 5m) = 36 meters east. Example 2: A diver leaps 1 meter straight up off a diving board, then falls downward for 5 meters before hitting the water. What is the diver’s displacement? The diver ended up 4 meters below the starting point, so her displacement is 4 meters downward, or -4 meters. (0 + 1 - 5 = -4). Even though the diver traveled six meters (one up, then five down), what matters is that the end point is four meters below the start point.

The car was displaced by (41m - 5m) = 36 meters east.

The diver ended up 4 meters below the starting point, so her displacement is 4 meters downward, or -4 meters. (0 + 1 - 5 = -4). Even though the diver traveled six meters (one up, then five down), what matters is that the end point is four meters below the start point.

Example 1 (cont. ): The problem tells us that the car took 8 seconds to go from the start point to the end point, so this is the change in time. Example 2 (cont. ): If the diver jumped at t = 7 seconds and hits the water at t = 8 seconds, the change in time = 8s - 7s = 1 second.

Example 1 (cont. ): The car changed its position by 36 meters over 8 seconds. vav=36m8s={\displaystyle v_{av}={\frac {36m}{8s}}=} 4. 5 m/s east. Example 2 (cont): The diver changed her position by -4 meters over 1 second. vav=−4m1s={\displaystyle v_{av}={\frac {-4m}{1s}}=} -4 m/s. (In one dimension, negative numbers are usually used to mean “down” or “left. " You could say “4 m/s downward” instead. )

Example 3: A man jogs for 3 meters east, then make a 90º turn and travels 4 meters north. What is his displacement? Draw a diagram and connect the start point and end point with a straight line. This is the hypotenuse of a triangle, so solve for its length of this line using properties of right triangles. In this case, the displacement is 5 meters northeast. At some point, your math teacher may require you to find the exact direction traveled (the angle above the horizontal). You can do this by using geometry or by adding vectors.

Draw a diagram and connect the start point and end point with a straight line. This is the hypotenuse of a triangle, so solve for its length of this line using properties of right triangles. In this case, the displacement is 5 meters northeast. At some point, your math teacher may require you to find the exact direction traveled (the angle above the horizontal). You can do this by using geometry or by adding vectors.

vf=vi+at{\displaystyle v_{f}=v_{i}+at}, or “final velocity = initial velocity + (acceleration * time)” Initial velocity vi{\displaystyle v_{i}} is sometimes written as v0{\displaystyle v_{0}} (“velocity at time 0”).

Example: A ship sailing north at 2 m/s accelerates north at a rate of 10 m/s2. How much did the ship’s velocity increase in the next 5 seconds? a = 10 m/s2 t = 5 s (a * t) = (10 m/s2 * 5 s) = 50 m/s increase in velocity.

a = 10 m/s2 t = 5 s (a * t) = (10 m/s2 * 5 s) = 50 m/s increase in velocity.

Example (cont): In this example, how fast is the ship traveling after 5 seconds? vf=vi+at{\displaystyle v_{f}=v_{i}+at} vi=2m/s{\displaystyle v_{i}=2m/s} at=50m/s{\displaystyle at=50m/s} vf=2m/s+50m/s=52m/s{\displaystyle v_{f}=2m/s+50m/s=52m/s}

vf=vi+at{\displaystyle v_{f}=v_{i}+at} vi=2m/s{\displaystyle v_{i}=2m/s} at=50m/s{\displaystyle at=50m/s} vf=2m/s+50m/s=52m/s{\displaystyle v_{f}=2m/s+50m/s=52m/s}

In our example, since the ship started going north and did not change direction, its final velocity is 52 m/s north.

“A train accelerates at 7 m/s2 for 4 seconds, and ends up traveling forward at a velocity of 35 m/s. What was its initial velocity?” vf=vi+at{\displaystyle v_{f}=v_{i}+at}35m/s=vi+(7m/s2)(4s){\displaystyle 35m/s=v_{i}+(7m/s^{2})(4s)}35m/s=vi+28m/s{\displaystyle 35m/s=v_{i}+28m/s}vi=35m/s−28m/s=7m/s{\displaystyle v_{i}=35m/s-28m/s=7m/s}

The circular velocity of an object is calculated by dividing the circumference of the circular path by the time period over which the object travels. When written as a formula, the equation is: v = (2πr) / T Note that 2πr equals the circumference of the circular path. r stands for “radius” T stands for “time period”

Example: Find the circular velocity of an object traveling a circular path with a radius of 8 m over a full time interval of 45 seconds. r = 8 m T = 45 s Circumference = 2πr = ~ (2)(3. 14)(8 m) = 50. 24 m

Example: v = (2πr) / T = 50. 24 m / 45 s = 1. 12 m/s The circular velocity of the object is 1. 12 m/s.