### Video Transcript

An object starts from rest and
accelerates along an eight-meter-long straight line. Its velocity reaches 12 meters per
second when it is at the end of the line. What is the object’s acceleration
along the line?

So, this question asks us to find
the acceleration of an object. We’re told that the object starts
from rest. In this question, the phrase “from
rest” means that the object is initially not moving. Therefore, its initial velocity is
zero meters per second. The question then tells us that the
object reaches a velocity of 12 meters per second after moving from rest along an
eight-meter-long straight line.

Recall that acceleration is defined
as the rate of change of velocity. This increase in velocity from zero
meters per second to 12 meters per second tells us that the object has accelerated
along the line. Now recall that velocity and
acceleration are vector quantities, meaning that they have a direction as well as a
magnitude.

The question tells us that the
object is moving along a straight line. This means that it doesn’t change
direction, so we know for sure that the acceleration’s direction is constant. Since we’re asked to calculate a
single value for this acceleration, we can assume that the magnitude of the
acceleration is constant too.

To help us answer the question,
let’s use symbols to rewrite the information that we’ve been given. We want to find the object’s
acceleration, and we can label this as 𝑎. At the moment, the value of 𝑎 is
unknown. Next, we can label the initial
velocity as 𝑢, which we’ve learnt is equal to zero meters per second. The final velocity can be labeled
as 𝑣, and this is equal to the 12 meters per second that the object reaches at the
end of the line. Lastly, we can label the distance
traveled along the line as 𝑠, and this is equal to eight meters.

Using this information, as well as
remembering that the object’s acceleration will be constant, we can use an equation
of motion, also known as a kinematic equation, to help us calculate the answer. The kinematic equation that we’ll
want to use is 𝑣 squared is equal to 𝑢 squared plus two times 𝑎 times 𝑠. We are solving for the acceleration
𝑎. This means we need to rearrange the
equation in order to make 𝑎 the subject.

First, we can subtract 𝑢 squared
from both sides. After canceling out the positive
and negative 𝑢 squared terms on the right-hand side, we are now left with 𝑣
squared minus 𝑢 squared equals two times 𝑎 times 𝑠. Let’s now divide both sides by two
𝑠. After canceling out two 𝑠 from the
numerator and denominator on the right-hand side, we are finally left with the
equation that we can use to find the acceleration. 𝑣 squared minus 𝑢 squared divided
by two 𝑠 is equal to the acceleration 𝑎.

We can now substitute into the
equation the values that we have for 𝑣, 𝑢, and 𝑠. This gives us that the acceleration
𝑎 is equal to the square of 12 meters per second minus the square of zero meters
per second all divided by two times eight meters.

Notice that the velocities in the
numerator each have units of meters per second. Taking the square of these units
gives units of meters squared per second squared. We can then cancel one factor of
meters from the terms in the numerator with the units of meters from the
denominator. This gives us overall units for the
acceleration of meters per second squared. When we type this expression into a
calculator, we get a value for this acceleration of nine meters per second
squared.

Therefore, our final answer is that
the object’s acceleration along the line is nine meters per second squared.