What is the relationship between time and distance covered by a body if it starts from rest under constant acceleration along a straight line?

In this explainer, we will learn how to use an object’s initial and final velocities and the object’s displacement to define acceleration by using the formula 𝑣=𝑢+2𝑎𝑠.

We can recall the definition of acceleration as follows.

The acceleration of an object is defined as the rate of change of that object’s velocity.

The average acceleration, 𝑎, of an object that changes its velocity by an amount Δ𝑣 over a time period Δ𝑡 is given by 𝑎=Δ𝑣Δ𝑡.

We can also recall that velocity is a vector quantity, which means that it has a direction as well as a magnitude. Since acceleration measures the rate of change of velocity, this means that acceleration is also a vector quantity. For motion along a straight line (in other words, an object that only moves back and forth along one axis), this effectively means that velocities and accelerations can be either positive or negative.

Our definition of acceleration is useful when we are interested in how an object’s velocity changes with time. But sometimes we are not so interested in how much time the acceleration occurs over, but rather in the distance traveled by the object during the process.

An example of when this might be the case is when considering the stopping distance of a car.

Let’s say we have a car that is traveling at some constant velocity when the driver notices an obstacle up ahead, as in the diagram below.

The driver needs to brake in order to bring the car to a halt before it reaches the position of this obstacle. Applying the brakes will cause the car to slow down; this is a negative acceleration or a deceleration. The important question is then not how much time it will take the car to come to a stop but how much distance it will cover before its velocity reaches zero. This way, the driver knows how far back from the obstacle they need to apply the brakes.

We will label this distance over which the velocity changes as 𝑠, the acceleration as 𝑎, the initial velocity as 𝑢, and the final velocity as 𝑣. Notice that here, and throughout this explainer, when we refer to “distance,” we mean the “the magnitude of the displacement.” In the case of motion along a straight line and no change of direction, these two terms are interchangeable. In the example of a stopping car, the final velocity, 𝑣, will be 0 m/s, but in general both the initial and final velocities may be nonzero.

We can derive an equation linking the acceleration and the distance as follows.

The distance 𝑠 traveled by an object moving with an average velocity, 𝑣av, over a time interval, Δ𝑡, is given by 𝑠=𝑣Δ𝑡.av

We can rewrite the average velocity, 𝑣av, as 𝑣=𝑣+𝑢2,av where 𝑣 is the final velocity and 𝑢 is the initial velocity.

We can use our definition of acceleration as 𝑎=Δ𝑣Δ𝑡 to rewrite Δ𝑡 as Δ𝑡=Δ𝑣𝑎=𝑣−𝑢𝑎.

Then, substituting in these expressions for 𝑣av and Δ𝑡 into our expression for the distance traveled, 𝑠, we get 𝑠=𝑣Δ𝑡=𝑣+𝑢2×𝑣−𝑢𝑎.av

Multiplying the terms on the top and the bottom of the fraction, we get 𝑠=𝑣−𝑢2𝑎.

Multiplying both sides by 2𝑎 and then adding 𝑢 to both sides, we can rewrite this equation as: 𝑣=𝑢+2𝑎𝑠.

The final velocity, 𝑣, the initial velocity, 𝑢, and the distance covered, 𝑠, for an object undergoing a uniform acceleration, 𝑎, are related by the formula 𝑣=𝑢+2𝑎𝑠.

Perhaps it seems strange that we are able to come up with a formula for the acceleration, 𝑎, that does not involve the quantity time when our definition of acceleration depends on the velocity change and the time interval over which that change occurs. The reason that it is possible to come up with a formula without using time is that time is related to displacement and velocity via 𝑣=Δ𝑠Δ𝑡.

There are a couple of restrictions on when we can use our formula:

1. The acceleration of the object must be constant.
2. The motion of the object must occur along a straight line.

Only when these two conditions are met may we use the formula 𝑣=𝑢+2𝑎𝑠.

Let’s have a look at an example in which we can apply this formula. In all of the examples within this explainer, we will use the same notation: the initial velocity is labeled 𝑢, the final velocity is labeled 𝑣, the acceleration is labeled 𝑎, and the distance over which this acceleration occurs is labeled 𝑠.

An object starts from rest and accelerates at 2 m/s2 along a 9 m long straight line. What final velocity does the object have?

Answer

Let’s begin by assigning labels to the values given to us in the question.

We are told that the object starts from rest, which means that its initial velocity is zero. So, we have that 𝑢=0/ms.

We are told that it accelerates over a 9 m long straight line, which means we have 𝑠=9m.

Finally, we are told that the acceleration is 2 m/s2, so we know that 𝑎=2/ms.

We can draw a sketch showing the situation.

In the sketch, we have labeled all of the quantities given to us in the question and also marked the final velocity, 𝑣, that we are trying to find. This final velocity is the velocity of the object after it has accelerated over the distance 𝑠=9m.

Since we are told the value of the acceleration is 𝑎=2/ms, we know this is a constant acceleration.

We are also told that the acceleration occurs over a straight line.

We can recall that since these two necessary criteria are met, our formula 𝑣=𝑢+2𝑎𝑠 may be used.

Looking at the formula, we see that we know the values for all of the quantities on the right-hand side: 𝑢=0/ms, 𝑎=9/ms, and 𝑠=9m.

Substituting in these values gives us an expression for the square of the final velocity 𝑣: 𝑣=(0/)+2×2/×(9).msmsm

Evaluating the right-hand side, we have 𝑣=36/.ms

The final step is to take the square root to give us 𝑣: 𝑣=√36/=6/.msms

So, we now have our answer. The final velocity, 𝑣, of the object is equal to 6 m/s. The direction of this velocity is the same direction as that of the acceleration.

Sometimes, we have situations where the formula 𝑣=𝑢=2𝑎𝑠 is helpful but where the quantity that we want to calculate is not the final velocity, 𝑣, of the object. In other words, perhaps we already know the value of this final velocity but want to find out the value of one of the other variables in the formula.

In this case, we need to rearrange the equation in order to make the subject of the equation into whichever quantity we would like to find the value of.

For example, if we were looking to calculate the value of 𝑎, it would be helpful to have an equation that reads 𝑎=…, and similarly for any of the other quantities.

Let’s have a look at an example in which we need to rearrange the formula.

An object starts from rest and accelerates along an 8 m long straight line. Its velocity reaches 12 m/s when it is at the end of the line. What is the object’s acceleration along the line?

Answer

Let’s begin by labeling the values given to us in the question.

We are told that the length of the straight line over which the object accelerates is 8 m, so we have that 𝑠=8m.

We are also told that the object accelerates from rest, meaning its initial velocity is 𝑢=0/ms, and that it reaches a velocity of 12 m/s at the end of the line, so we have that the final velocity is 𝑣=12/ms.

We can draw a sketch showing this information.

We have labeled all of the quantities given to us in the question in our sketch and also marked the acceleration, 𝑎, that we are asked to find.

We can recall our formula: 𝑣=𝑢+2𝑎𝑠.

In this case, the quantity that we are trying to work out is the acceleration, 𝑎. This means that we need to rearrange the formula to make 𝑎 the subject.

We begin by subtracting 𝑢 from both sides of the formula, which gives us 𝑣−𝑢=2𝑎𝑠.

Then, we divide both sides by 2𝑠, giving us 𝑣−𝑢2𝑠=𝑎.

Finally, swapping the left-hand and right-hand sides over gives us 𝑎=𝑣−𝑢2𝑠.

Now that we have made the acceleration, 𝑎, the subject, we are ready to substitute in our values: 𝑣=12/ms, 𝑢=0/ms, and 𝑠=8m.

When we do this, we get the following expression for 𝑎: 𝑎=(12/)−(0/)2×8.msmsm

Evaluating the expressions in the numerator and the denominator on the right-hand side, we get 𝑎=144/16.msm

When we do this division on the right-hand side, we get that 𝑎=9/.ms

And so our answer is that the acceleration of the object along the line is 9 m/s2.

In both the examples we have seen so far, the object has started from rest and the acceleration has acted to increase the velocity of the object from an initial velocity of 0 m/s to a nonzero final velocity.

In general, the initial velocity of the object may take any value. In other words, the object may, in general, already be moving before the period of acceleration.

Recalling that velocity and acceleration are both vector quantities, we can see that for motion along a straight line, there are two possibilities:

1. The acceleration acts in the same direction as the velocity and acts to increase the velocity in this direction.
2. The acceleration acts in the opposite direction to the velocity and acts to decrease the magnitude of the velocity.

These two cases are illustrated visually in the diagram below.

Let’s look at an example in which the initial velocity is not zero.

An object has an initial velocity that increases to 14 m/s as the object accelerates at 5 m/s2 in the direction of its velocity. The object accelerates along a 17.1 m long straight line. What is the object’s initial velocity?

Answer

Let’s begin by labeling the quantities given to us in the question.

We are told that the velocity of the object increases to 14 m/s while it accelerates in the direction of its velocity.

This means that both the velocity and the acceleration have the same sign.

We have that the final velocity is 𝑣=14/ms. Since we are told that the acceleration is 5 m/s2, we have that 𝑎=5/ms.

We are told that the object accelerates over a 17.1 m long straight line, so we have 𝑠=17.1m.

The question asks us to find the initial velocity, which we will label 𝑢.

We can draw a sketch showing this information as follows.

We have a constant value of acceleration and we are told that the motion occurs over a straight line, so our formula 𝑣=𝑢+2𝑎𝑠 may be used here.

Since we are asked to find the value of the initial velocity, let’s rearrange the formula to make 𝑢 the subject.

Subtracting 2𝑎𝑠 from both sides, we get 𝑣−2𝑎𝑠=𝑢.

Then, swapping the left-hand and right-hand sides of the equation over, we have 𝑢=𝑣−2𝑎𝑠.

Now we are ready to substitute in the values from the question: 𝑣=14/ms, 𝑎=5/ms, and 𝑠=17.1m.

Doing this gives us the following expression for the square of the initial velocity: 𝑢=(14/)−2×5/×(17.1).msmsm

Evaluating the right-hand side gives 𝑢=196/−171/𝑢=25/.msmsms

Finally, taking the square root of both sides gives us 𝑢=5/.ms

And so, we have found that the initial velocity of the object was 5 m/s. Since this value is positive, it is in the same direction as both the acceleration and the final velocity.

In this last example, the acceleration was in the same direction as the initial velocity of the object, so the effect of this acceleration was to increase the velocity of the object in the direction it was already moving.

Now, we will look at an example in which the acceleration acts in the opposite direction to the initial velocity.

An object has an initial velocity that decreases to 10 m/s as the object accelerates in the opposite direction to its velocity. The object moves along a 60 m long straight line accelerating with a magnitude of 6.5 m/s2. What is the object’s initial velocity to the nearest metre per second?

Answer

Let’s begin by assigning labels to the values given to us in the question.

We are told that the velocity decreases to a final value of 10 m/s, so we have 𝑣=10/ms.

We are also told that the object moves along a 60 m long straight line, so 𝑠=60m.

Finally, we are told that the object accelerates in the opposite direction to its velocity, with a magnitude of 6.5 m/s2. Since we have chosen to take the velocity 𝑣 to be positive and the acceleration is in the opposite direction to this, then this acceleration must be negative. So, we have 𝑎=−6.5/ms.

To make these directions clear, we can draw a sketch of the situation as follows.

In this sketch, we have labeled the quantities we know the values of, along with their directions. We have also labeled the initial velocity, 𝑢, also with its direction; the value of 𝑢 is what we are asked to find. Note that, in this sketch, we have not included the negative sign on the acceleration because this is implicitly included by the direction of the acceleration arrow; the value of the acceleration is +6.5 m/s2 in the direction that this arrow is pointing.

We have a constant value for the acceleration, 𝑎, and we are told that the motion is along a straight line. This means that we are in a position to use our formula: 𝑣=𝑢+2𝑎𝑠.

Since we are asked to find 𝑢, let’s rearrange this formula to make 𝑢 the subject.

Subtracting 2𝑎𝑠 from each side, we have 𝑣−2𝑎𝑠=𝑢.

Swapping the left-hand and right-hand sides over, we get 𝑢=𝑣−2𝑎𝑠.

We are now ready to substitute in our values into this equation. When we do this we need to take care with the signs of all of the quantities.

Substituting in 𝑣=10/ms, 𝑎=−6.5m, and 𝑠=60m, we get the following expression for the square of the initial velocity: 𝑢=(10/)−2×−6.5/×(60).msmsm

Evaluating the right-hand side (and taking care with the negative signs), we get 𝑢=100/−−780/𝑢=100/+780/𝑢=880/.msmsmsmsms

Then, taking the square root of both sides gives us 𝑢=29.66…/.ms

Here, the ellipsis indicates that there are further decimal places.

Finally, we note that the question asked for our answer to the nearest metre per second. Rounding our result to the nearest metre per second gives 𝑢=30/.ms

And so we have found that the initial velocity of the object, to the nearest metre per second, is equal to 30 m/s.

Perhaps the most useful rearrangement of the formula 𝑣=𝑢+2𝑎𝑠 is one that we have not come across so far: the rearrangement making 𝑠 the subject.

This is the form we would need this formula in for calculating the stopping distance of a car, which was our initial motivation for wanting an equation linking acceleration and displacement.

To make 𝑠 the subject, we first subtract 𝑢 from both sides of the equation, giving us 𝑣−𝑢=2𝑎𝑠.

Then, dividing through by 2𝑎 and swapping the left-hand and right-hand sides of the equation over, we have 𝑠=𝑣−𝑢2𝑎.

Let’s think back to the case of a car in which the driver notices an obstacle ahead and so applies the brakes to bring the car to a stop. As a reminder, the situation is shown in the diagram below.

We know the final velocity of the car will be 𝑣=0/ms. If we were to know the magnitude of the acceleration, 𝑎, that the brakes provide and the initial velocity, 𝑢, that the car was traveling with, we would be able to use the equation 𝑠=𝑣−𝑢2𝑎 to calculate the distance, 𝑠, that the car would travel before it stops.

We will finish by working through an example problem in which the goal is to calculate the distance traveled.

A large bird has to run while flapping its wings to launch itself into the air. The bird needs to be running at 5.745 m/s to start to fly. If the bird can accelerate at 1.65 m/s2, how far must it run before it can take off? Give your answer to one decimal place.

Answer

We will begin by labeling the quantities that we are given.

The question asks us to work out the distance, 𝑠 that the bird must run before it can take off.

We are told that the bird needs to be running at 5.745 m/s in order to start to fly, so we have that the final velocity is 𝑣=5.745/ms. We can also assume that the bird is starting from rest, giving us an initial velocity of 𝑢=0/ms. The final bit of information we are given is that the bird can accelerate at 𝑎=1.65/ms.

We can draw a sketch showing this information as follows.

The rate of acceleration given to us is a constant value. We can assume that, as shown in our sketch, the bird runs in a straight line while preparing to take flight.

In this case, we can use our formula: 𝑣=𝑢+2𝑎𝑠.

Since we are trying to find the value of 𝑠, we need to make 𝑠 the subject.

So, we first subtract 𝑢 from each side: 𝑣−𝑢=2𝑎𝑠.

Then, we divide both sides by 2𝑎, and swap the left-hand and right-hand sides over: 𝑠=𝑣−𝑢2𝑎.

We are now in a position to substitute in our values 𝑣=5.745/ms, 𝑢=0/ms, and 𝑎=1.65/ms. Doing this gives us the following expression for 𝑠: 𝑠=(5.745/)−(0/)2×1.65/.msmsms

Evaluating the expressions in the numerator and the denominator on the right-hand side gives 𝑠=33.005025/3.3/.msms

Doing the division gives us 𝑠=10.00152….m

Here, the ellipsis indicates that there are further decimal places.

The final step is to note that the question asks for our answer to one decimal place. Rounding our result to one decimal place gives 𝑠=10.0.m

And so, we have found that the bird must run 10.0 m before it can take off.

• The final velocity, 𝑣, the initial velocity, 𝑢, and the distance covered, 𝑠, for an object undergoing a uniform acceleration, 𝑎, are related by the formula 𝑣=𝑢+2𝑎𝑠.
• This formula only applies when the acceleration, 𝑎, is constant and the motion is in a straight line.
• Acceleration is a vector quantity. For motion along a straight line, it can either be positive (in the same direction as the initial velocity) or negative (in the opposite direction to the initial velocity).
• We can use the formula 𝑣=𝑢+2𝑎𝑠 when any three of the quantities in this formula are known and we are trying to find the fourth. If the quantity we are trying to find is not the final velocity, 𝑣, then we must begin by rearranging the formula to make the quantity we are trying to find the subject.