How to find distance with velocity and time and acceleration

Índice Show

Final Velocity
Distance Traveled
Acceleration
Example - Accelerating Motorcycle

Updated December 08, 2020

By Lee Johnson

Kinematics is the branch of physics that describes the basics of motion, and you’re often tasked with finding one quantity given knowledge of a couple of others. Learning the constant acceleration equations sets you up perfectly for this type of problem, and if you have to find acceleration but only have a starting and final velocity, along with the distance travelled, you can determine the acceleration. You only need the right one of the four equations and a little bit of algebra to find the expression you need.

The acceleration formula applies to constant acceleration only, and a stands for acceleration, v means final velocity, u means starting velocity and s is the distance travelled between the starting and final velocity.

There are four main constant acceleration equations that you’ll need to solve all problems like this. They’re only valid when the acceleration is “constant,” so when something is accelerating at a consistent rate rather than accelerating faster and faster as time goes on. Acceleration due to gravity can be used as an example of constant acceleration, but problems often specify when the acceleration continues at a constant rate.

The constant acceleration equations use the following symbols: a stands for acceleration, v means final velocity, u means starting velocity, s means displacement (i.e. distance traveled) and t means time. The equations state:

v=u+at\\ s=0.5(u+v)t\\ s=ut+0.5at^2\\ v^2=u^2+2as

Different equations are useful for different situations, but if you only have the velocities v and u, along with distance s, the last equation perfectly meets your needs.

Get the equation in the correct form by re-arranging. Remember, you can re-arrange equations however you like provided you do the same thing to both sides of the equation in every step.

v^2=u^2+2as

Subtract u2 from both sides to get:

v^2-u^2=2as

Divide both sides by 2 s (and reverse the equation) to get:

a=\frac{v^2-u^2}{2s}

This tells you how to find acceleration with velocity and distance. Remember, though, that this only applies to constant acceleration in one direction. Things get a bit more complicated if you have to add a second or third dimension to the motion, but essentially you create one of these equations for motion in each direction individually. For a varying acceleration, there is no simple equation like this to use and you have to use calculus to solve the problem.

Imagine a car travels with constant acceleration, with a velocity of 10 meters per second (m / s) at the start of a 1 kilometer (i.e. 1,000 meter) long track, and a velocity of 50 m / s by the end of the track. What is the constant acceleration of the car? Use the equation from the last section, remembering that v is the final velocity and u is the starting velocity. So, you have v = 50 m/s, u = 10 m / s and s = 1000 m. Insert these into the equation to get:

a=\frac{50^2-10^2}{2\times 1000}=\frac{2400}{2000}=1.2\text{ m/s}^2

So the car accelerates at 1.2 meters per second per second during its journey across the track, or in other words, it gains 1.2 meters per second of speed every second.

Velocity is the change in position (x), or distance, over time. If you know the change in position and the amount of time taken to complete the journey, you can determine velocity. Similarly, if you have any two of these variables, you can always solve for the third.

The relationship between these three variables is as follows:

V = \dfrac{\bigtriangleup x}{t}

A car drives from Baltimore to Washington, D.C. in 1.5 hours. If you know that it's 38 miles between the two cities, what was the car's average velocity during the trip? Since, this is a trip that goes in one direction, the change in position is the same as distance. Since you know time and the distance, you can solve for velocity by plugging in the distance formula in physics:

V = \dfrac{\bigtriangleup x}{t}

V = \dfrac{38}{1.5} = 25.3

So you know your answer is 25.3, but this isn't quite complete: 25.3 what? Units are just as important as the numerical answer when it comes to physics problems, so don't lose track of what you're using to measure distance and time. Since you're measuring distance in miles and time in hours, your final answer is miles divided by hours, or miles per hour.

A bicyclist completes a 550 meter race in 1.5 hours. What is the bike's velocity in meters per second? Here, since you need to determine the velocity in meters per second, first convert time to seconds:

(1.5 hours)(60 minutes)(60 seconds) = 5,400 seconds

Then, plug your known variables into the velocity formula:

V = \dfrac{\bigtriangleup x}{t}

V = \dfrac{550}{5400} = 0.1m/s

If you know how fast and how long something was traveling, you can solve for the distance traveled. You just need to rearrange the velocity formula above to get the distance formula in physics:

{\bigtriangleup x} =(velocity)(time)

A plane travels 150 miles per hour on it's way from Atlanta to San Diego. How far has the plane traveled in 3.5 hours?

Since the plane appears to be going in one direction (toward San Diego) in a straight line, you can assume that the change in position equals distance. Plug your known variables into the distance formula:

{\bigtriangleup x} =(150mph)(3.5h) = 525 miles

Tips

Make sure to pay attention to units when using the distance formula in physics. If you're using a velocity that's miles per hour, and you're solving for distance, make sure your time is in hours too.

If you need to solve for time, you just rearrange the formula one more time:

time = \dfrac{\bigtriangleup x}{velocity}

Say a turtle crawls at 3 mph. How long will it take the turtle to finish a 5-mile race?

time = \dfrac{5}{3} = 1.67 h

People tend to use "speed" for "velocity" and vice versa, but they are slightly different concepts. Speed doesn't take into account direction, while velocity does. If you look at the formula, velocity is the change in position over time, while speed is distance over time. Let's look at an example to illustrate:

Say you drive 20 miles from your house to your college campus and then head back. It took you an hour round trip. What is your average speed?

You know your total distance and the time taken, so plug into the formula for speed:

speed= \dfrac{distance}{time}

speed= \dfrac{40}{1} = 40mph

Now, what is your average velocity? Keep in mind that you use change in location or displacement to determine velocity because direction matters:

velocity= \dfrac{\bigtriangleup distance}{time}

Since you end at your beginning location, your change in position or distance is actually 0, which means your velocity is also 0. Velocity is equal to the formula for speed only if you're traveling in a straight line.

va = (v1 + v0) / 2 (1)

where

va = average velocity (m/s)

v0 = initial velocity (m/s)

v1 = final velocity (m/s)