# Lesson 4: Acceleration

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Download here: Ontario Curriculum Expectations

The expectations of life depend upon diligence; the mechanic that would perfect his work must first sharpen his tools.

Confucius

#### Equations Have Limits

No equation is perfect. With all of the equations that you will be looking at in this activity, you must be aware of the specific situations, and conditions, in which they can, and cannot, be used. Understanding these conditions will help you to determine which equation is best suited to a given problem and how to use it to accurately calculate the required value. The following three conditions apply to all of the equations you will learn in this activity:

- The acceleration must be constant.

It does not matter if the acceleration is: positive, where the object is speeding up; negative, where the object is slowing down; or even zero, where the object is maintaining a constant speed. In each of these cases, the acceleration need not change in order for the equation you will be using to be valid. For any given problem, there should be just one value for acceleration. If an object undergoes two different accelerations within the same problem, these two separate motions must be calculated separately.

- The units must be consistent. Except for very rare situations the following units should be used for all quantities:

Displacement = metres (m)

Velocity = metres per second (m/s)

Time = seconds (s)

Acceleration = metres per second squared (m/s^{2})

Using units like kilometres (km) or kilometres per hour (km/h) will often lead to unnecessarily awkward units for your answers.

- All motion must be in the same direction.

Displacement, velocity, and acceleration are all vectors and, therefore, all have direction. When using these equations, you must ensure that the direction of all of the quantities is the same. Sometimes, when the motion is in different directions but in the same dimension (east and west, north and south, left and right) it is easy to do. For example, two velocities in opposite directions, v_{1} = 10 m/s [S] and v_{2} = 30 m/s [N], could be simplified v_{1} to -10 m/s [N]. Conversely, when there is more than one dimension (north and east, left and up) the calculation of the resulting velocity becomes much more complicated.

#### Enrichment

There is a very interesting, and brief, discussion that addresses some of the above topics in the first four paragraphs at The Physics Hypertextbook website.

#### Starting with what we know…

You have already worked with the equation for velocity, . However, you will now be examining the motion of objects that are accelerating, which means that the object’s velocity must be changing. It no longer makes sense to refer to just one velocity. However, we may be interested in finding the average velocity, as we did in activity two, for an object over a period of time. To indicate that it is an average velocity, we use the following equation:

By using the term “average” we recognize that at times the velocity may be faster or slower, but overall, we can still get a general idea of how quickly it is moving.

By rearranging equation 1, we get an equation for calculating displacement:

#### …and moving to something new.

Think about how you would determine your average mark in a course if you earned 75% on the first test and an 87% on the second. You would just add up the two marks and divide by two. Written mathematically, we would get:

Your average mark would be 81%. We can take a similar approach to calculating the average velocity. We just find the average of the starting velocity and the ending velocity:

We are only able to do this in cases where velocity of the object changes at a constant rate; that is, where rate of acceleration is constant. If the rate of acceleration is not constant then the equation would not be valid.

#### Practice Questions

A helicopter flying at 20 m/s [E] undergoes a constant acceleration to a final velocity of 30 m/s [W]. What is the helicopter’s average velocity?

**Answer**

Given:

= 20 m/s [E]

= 30 m/s [W]

We need all of the values to be in the same direction. Let’s make west positive and east negative.

Therefore,

= 20 m/s [E] = – 20 m/s [W]

Required:

= ?

Solution:

The helicopter’s average velocity is 5 m/s [W].

Acceleration is the rate at which the velocity changes. Written out mathematically we get: . However, this can be easily be rearranged in terms of to get:

#### Practice Questions

- A car is travelling at 14 m/s [E]. It accelerates at 3.5 m/s
^{2}[E] for 8.0 s. What is the car’s final velocity?

**Answer**

Given:

= 14 m/s [E]

= 3.5 m/s^{2} [E]

Δt = 8.0 s

Required:

= ?

Solution:

The car’s final velocity is 42 m/s [E].

#### Two more motion equations…

By combining equations 2, 3, and 4, we can derive a new equation for displacement. First we need to substitute equation 4 into equation 3 and simplify:

Then we need to substitute this into equation 2 and simplify:

#### Practice Questions

- While riding your bike at 8.0 m/s, you start to accelerate down a steep hill at 3.0 m/s
^{2}. How far have you travelled in 12.0 s?

**Answer**

Given:

= 8.0 m/s

= 3 m/s^{2}

Δt = 12.0 s

In this case, even though no directions have been given, the values are still vectors. In all parts of the bike ride, the rider (you in this case) is moving forward. Therefore, the direction is forward.

Required:

= ?

Solution:

You will have travelled 312 m down the hill.

We can derive another equation for displacement by substituting equation 3 into equation 2:

#### Practice Questions

- A train enters a tunnel at a speed of 10.0 m/s and exits 12.0 seconds later at a speed of 18.0 m/s. How long is the tunnel? What is the train’s acceleration while in the tunnel?

**Answer**

Given:

= 10.0 m/s

= 18.0 m/s

Δt = 12.0 s

The direction for all values is forward.

Required:

= ?

= ?

Solution:

= 168 m

= 0.667 m/s^{2}

The train accelerates at 0.667 m/s^{2} while travelling through a 168 m long tunnel.