Interactive textbook: Acceleration

Acceleration

Acceleration is defined as the rate of change of velocity:

a c c e l e r a t i o n space equals space fraction numerator c h a n g e space i n space v e l o c i t y over denominator t i m e end fraction

If the velocity of a body changes then it has acceleration, regardless of whether it is slowing down or speeding up. This differs from everyday language where acceleration means getting faster and deceleration is getting slower. 

If a moving body changes direction, its velocity changes. This means the body accelerates even if the speed is constant.

Acceleration is a vector quantity. Its direction is in the direction of the change in velocity. The acceleration vector can be found by subtracting the vector representing the initial velocity from the vector representing the final velocity. To subtract a vector, we reverse the direction of the vector and add it:

Now consider a body travelling with an increasing speed around a corner:

The black arrows represent the velocity vectors and the red ones the change of velocity. We can see that the change in velocity itself also changes direction. This means that the acceleration has changed direction.

 

A body travels around a circle at constant speed.

The velocity and acceleration are:

The magnitude of both is constant but the direction changes.

Constant acceleration

The motion of a car travelling between two towns is very complicated. The direction, speed and acceleration will be changing all the time with every corner, junction and interaction with other road users. This would be impossible for an IB physicist to model so we choose instead only to consider the case of constant acceleration (and, hence, in a straight line).

If the motion is along a horizontal line, there are only two possible directions: left and right. The sign of the vectors that represent the motion can only have two directions. It makes sense to define right as positive and left as negative, like the \(x\)-axis of a graph. 

The signs of the vectors give the direction of the quantity:

  • The sign of displacement tells us whether the body is to the left or right of the point we decide to be zero.
  • The sign of velocity tells us the direction of motion.
  • The sign of acceleration tells us the direction of change of velocity.

For example, a body with negative displacement, positive velocity and negative acceleration would be to the left of zero, moving to the right and slowing down. A body with negative velocity and negative acceleration would be moving left and getting faster.

TOK: Intuition is a powerful thing and it is difficult to go against it. This isn't the only time you will come across this problem.

A body moves to the left with reducing velocity.

The acceleration is:

When solving problems you can choose, usually we take right as positive in which case the answer would be positive.

If the acceleration is constant then it can be defined by the following equation:

\(a={v-u\over t}\)

  • \(a\) is acceleration in \(\text{m s}^{-2}\)
  • \(v\) is final velocity in \(\text{m s}^{-1}\)
  • \(u\) is initial velocity in \(\text{m s}^{-1}\)
  • \(t\) is time in \(\text{s}\)

Let's consider an example:

\(a={v-u\over t}={-15-(-5)\over 5}=-2\text{ m s}^{-2}\)

The acceleration is negative, which means it is towards the left.

We should never forget that displacement, velocity and acceleration are vectors but in one dimension they add in the same way as positive and negative numbers. We don't have to bother with the arrows and so won't use vector notation when dealing with examples of constant acceleration; the sign is all we need to be able to know their direction.

Time can also be positive and negative but time is a scalar so the sign has nothing to do with direction, negative time represents a time before the clock was started.

A body travelling at 10 m s-1 undergoes a constant acceleration of -5 m s-2.

What is the velocity after 4 s?

a = (v - u)/t

v = u + at = 10 - 4 x 5

The 'suvat' equations of constant acceleration

'Suvat' is not a word, but its letters just represent the different quantities for motion with constant acceleration:

  • \(s\) is displacement in \(\text{m}\)
  • \(u\) is initial velocity in \(\text{m s}^{-1}\)
  • \(v\) is final velocity in \(\text{m s}^{-1}\)
  • \(a\) is acceleration in \(\text{m s}^{-2}\)
  • \(t\) is time in \(\text{s}\)

These letters are used to write equations that relate the different quantities. From the definition of acceleration, we already have:

\(a={v-u\over t} \Rightarrow v=u+at\) (1)

We also know that the average velocity is given by:

\(v_\text{av}={s\over t}\)

If the acceleration is constant:

 \(v_\text{av}={v+u\over 2}\)

It might seem strange that the average is found by halving the sum of the maximum and minimum values. This works because the acceleration is constant; the velocity increases at a constant rate like the heights of the following students:

Combining these two equations gives an equation for the displacement in terms of the final and initial velocity:

\(v_\text{av}={s\over t}={v+u\over 2}\Rightarrow s={1\over2}(v+u)t\) (2)

We can use these two equations to solve any problem involving constant acceleration, but sometimes two steps would be required. Combining these equations each time substituting for another quantity gives us four single-step equations.

Substituting \(v=u+at\) into \(s={1\over2}(v+u)t\) gives:

\(s={1\over 2}(u+at+u)t\)

\(\Rightarrow s=ut+{1\over2}at^2\) (3)

Rearranging the first equation and substituting \(t={v-u\over a}\) into \(s={1\over2}(v+u)t\)  gives:

\(s={(v+u)\over 2}{(v-u)\over a}\)

\(\Rightarrow v^2=u^2+2as\) (4)

Solving problems

Use these equations tactically, considering the three pieces of information that the question provides and the quantity that you are looking for.

  • Been given initial velocity, acceleration and displacement and need to find final velocity? Use equation 4.
  • What if you have initial velocity, acceleration and displacement and are asked to calculate time? Use equation 3.

What real-world situations can you imagine exam questions being based on?

A body is thrown upwards from the ground with a velocity of 20 m s-1.

Which equation would you use to calculate the time taken to return to the ground?

We know

u = 20 m s-1

s = 0 m

a = -10 m s-2

and we want to find t

Total Score:

All materials on this website are for the exclusive use of teachers and students at subscribing schools for the period of their subscription. Any unauthorised copying or posting of materials on other websites is an infringement of our copyright and could result in your account being blocked and legal action being taken against you.