P5 K) Acceleration – Part 2
Uniform acceleration means constant acceleration. An example of uniform acceleration is acceleration due to gravity, which is roughly 9.8 m/s2. The acceleration due to gravity is the same value as gravitational field strength (9.8 N/kg).
We are now going to have a look at another formula that we can use to work out the acceleration of an object. This formula is shown below.
Like the formula in the previous section, u is the initial velocity and v is the final velocity (both in m/s). a is acceleration (in m/s2) and s is distance (in m).
Sometimes you will see this formula rearranged to give us:
Also, sometimes you will see “x” instead of “s” in this formula to represent distance; it is totally up to you whether you have x or s in the formula.
A car accelerates from rest with a uniform acceleration of 5 m/s2. The car has a final velocity of 8 m/s. Find the distance that the car travels during this time.
The question is asking us to find the distance that the car has travelled, which is the value of s in the formula below.
The question tells us that the car starts from rest, so the initial velocity is 0 m/s (u = 0). We are also told that the final velocity is 8 m/s (v = 8) and the acceleration rate is 5 m/s2 (a = 5). We sub all of these values into the formula and solve to find the value of s in the same way that we would solve an algebraic maths equation.
We want to find the value of s and not 10s. Therefore, we divide both sides of the equation by 10.
This tells us that the distance is 6.4 metres.
A car is travelling at 15 m/s. It then accelerates uniformly at a rate of 3 m/s2. The car covers 49 metres whilst it is accelerating. Work out the final velocity of the car to one decimal place.
We are looking for the final velocity of the car, which is the value of v. Therefore, it makes sense to use the rearranged formula with v2 as the subject; this formula is shown below.
The question tells us that the initial velocity of the car is 15 m/s (u = 15), the acceleration rate is 3 m/s2 (a = 3) and the distance travelled is 49 metres (s = 49). We sub these values into the formula.
We want to find the value of v and not v2. Therefore, we square root both sides of the equation.
The question asks us to round our answer to 1 decimal place, which means that the final velocity of the car is 22.8 m/s.
A car slows down uniformly before it passes a school. The rate of deceleration is 1.5 m/s2 and whilst decelerating, the car covers 48 metres. After decelerating, the car has a velocity of 5 m/s. What is the initial velocity of the car?
I am going to use the formula with v2 as the subject; this formula is shown below.
The question is asking us to find the initial velocity, which is the value of u. We are told in the question that the final velocity is 5 m/s (v = 5) and the distance travelled is 48 metres (s = 48). The car is decelerating, and we know that deceleration is negative acceleration. Therefore, the acceleration is -1.5 m/s2 (a = -1.5). We sub all of these values into the formula and solve to find the value of u.
The first step in finding the value of u is to get u2 by itself; we need to move the -144 from the right side of the equation to the left. We are able to do this by doing the opposite, and the opposite of taking 144 is adding 144; so we add 144 to both sides of the equation.
We want to find the value of u and not u2. Therefore, we square root both sides of the equation.
Therefore, the initial velocity of the car is 13 m/s.
Newton’s Second Law
We can also use Newton’s second law to work out acceleration. Newton’s second law is shown below:
Force is the resultant force measured in Newtons (N), mass is measured in kilograms (kg) and acceleration is measured in metres per second squared (m/s2).
We will be looking at Newton’s second law in more detail in a later section; click here to be taken to this section