Video Transcript
An object has an initial velocity
of three metres per second and accelerates in the direction of its velocity along a
straight line at a rate of four metres per second squared. Its velocity reaches 11 metres per
second when it is at the end of the line. What is the length of the line?
Okay, so in this question, we’ve
got an object. Let’s say it starts here. And let’s say it’s moving in this
direction to the right with its initial velocity which we’ve been told is three
metres per second. Now, we’ve been told that it
accelerates in the direction of its velocity along a straight line at a rate of four
metres per second squared. So the object is travelling in a
straight line towards the right, as we’ve drawn it. And by the time it gets to the end
of the line, we’ve been told that it’s moving at 11 metres per second. And we’ve been asked to find the
length of this line, from the start position of the object to the finish
position.
Let’s say that the length of this
line is 𝑠. And 𝑠 is what we’re trying to
find. As well as this, let’s label the
initial velocity of the object as 𝑢, three metres per second. Let’s say the final velocity of the
object is 𝑣, 11 metres per second. And let’s say that the acceleration
of the object is 𝑎, four metres per second squared. Now, the equation that links 𝑢,
𝑣, 𝑎, and 𝑠 is this one here: 𝑣 squared is equal to 𝑢 squared plus two
𝑎𝑠. Or, the final velocity of the
object squared is equal to its initial velocity squared plus two times its
acceleration times the distance it travels.
Now, it’s important to know that
for this equation to apply, the acceleration must be a constant value. And in this case, we’ve been told
that it’s four metres per second squared which is a constant. As well as this, the distance that
the object travels in order for this equation to apply must be in a straight
line. And this is the case as we’ve been
told in the question. Therefore, we can use this equation
to solve our problem. We already know the values of 𝑣,
𝑢, and 𝑎. All we need to do is to rearrange
to solve for 𝑠. We can start by subtracting 𝑢
squared from both sides of the equation. This way, it cancels on the
right-hand side. And what we’re left with is 𝑣
squared minus 𝑢 squared on the left and two 𝑎𝑠 on the right. Then, we can divide both sides of
the equation by two 𝑎. This way the two on the right
cancels and the 𝑎 on the right cancels. And we have 𝑣 squared minus 𝑢
squared divided by two 𝑎 is equal to 𝑠.
So at this point, all we need to do
is to plug in the values. 𝑠 is equal to 𝑣, which is the
final velocity, squared, so 11 metres per second squared, minus 𝑢, which is the
initial velocity, squared, which is three metres per second whole squared, divided
by two times the acceleration, which is four metres per second squared.
Expanding the parentheses in the
numerator, we have 121 metres squared per second squared minus nine metres squared
per second squared. And so, the numerator becomes 112
metres squared per second squared. Then, we divide this by two times
four, which is eight, metres per second squared. And so, the per second squared in
the numerator and the denominator cancel. And one factor of metres in the
numerator cancels with the one factor of metres in the denominator. Overall, we’re just left with a
unit of metres, which is good because we’re calculating a distance which has units
of metres. And hence, we’re left with the
final answer to our question. The length of the line along which
the object moves is 14 metres.
