### Video Transcript

In 1967, New Zealander Burt Munro
set the world record for the Indian motorcycle on the Bonneville Salt Flats of Utah
of 295.38 kilometers per hour. The one-way course was 8.00
kilometers long. Acceleration rates are often
described by the time it takes to reach 96.0 kilometers per hour from rest. If this time was 4.00 seconds and
Burt accelerated that this rate until he reached his maximum speed, how long did it
take Burt to complete the course?

Let’s begin by highlighting some of
the vital information we’re given. We’re told that Burt’s maximum
speed was 295.38 kilometers per hour. The course he travelled was 8.00
kilometers long. We’re told that it took Burt 4.00
seconds in order to reach a speed of 96.0 kilometers per hour from rest. We want to solve for the total time
it took Burt to travel a course. We’ll call that 𝑡.

Let’s begin by solving for Burt’s
acceleration over the first four seconds of travel. Recall that acceleration is defined
as the change in velocity divided by the change in time. In our case, we’re told that Burt
was able to achieve a speed of 96.0 kilometers per hour in a time of 4.00
seconds. Before we calculate this
acceleration, we want to convert the units of kilometers per hour to meters per
second.

To do that, let’s recall that one
kilometer is equal to 1000 meters and one hour is equal to 3600 seconds. So if we take our given speed of
96.0 kilometers per hour and multiply it by 1000 meters per 3600 seconds, then this
multiplication will give us a result in units of meters per second. This is equal to 26.67 meters per
second. Now we can insert this value for
speed in meters per second into the numerator of our acceleration equation.

And when we calculate this
fraction, we find a value of six and two-thirds meters per second squared, which we
can abbreviate as 6.667. This is the acceleration that Burt
experienced up until reaching the maximum speed, 𝑣 sub max. Since we want to solve for the time
it took Burt to complete the whole 8.00 kilometer course, let’s find out how long it
took him to reach a speed of 𝑣 max accelerating from rest at this rate.

Since we have a constant
acceleration value, let’s refer to the kinematic equations to see if they can be
useful to us. Remember that the kinematic
equations are true whenever acceleration is constant, as it is in our case. Looking at the very first kinematic
equation listed, we see it can be useful to us because 𝑣 sub 𝑓 in our case is 𝑣
sub max; 𝑣 sub zero is zero because Burt started from rest; 𝑎, we just solved for;
and 𝑡 is what we want to find.

Since we don’t know whether the 𝑡
that were solving for is the total time it took Burt to complete the course, let’s
provisionally call it 𝑡 sub one. If we divide both sides of this
equation by 𝑎, then the acceleration term on the right cancels out and we find that
𝑡 sub one is equal to the 𝑣 max divided by 𝑎. We know 𝑎; we solved for that as
6.667 meters per second squared. And 𝑣 max is given to us as 295.38
kilometers per hour.

Before we compute this fraction,
like before we’ll want to convert a speed in kilometers per hour to a speed in
meters per second. We’ll take the same approach as
before, by multiplying this fraction by 1000 meters per 3600 seconds. The product of this multiplication
is 82.05 meters per second. Now we take that value and
substitute it in for the 𝑣 max value given in units of kilometers per hour. Performing this division, we find
that 𝑡 one is equal to 12.308 seconds. Knowing that, here is what we want
to figure out. If 𝑡 one equals 12.308 seconds,
then how far would Burt travel in that time?

Would it be equal to the length of
the course 8.00 kilometers? If it is, then we’ve already solved
for 𝑡. If not, we’ll have to take another
step. Let’s refer back to our list of
kinematic equations to solve for the distance that Burt will travel in a time of 𝑡
one. The third kinematic equation from
the top will be useful in answering this question. It tells us that the distance Burt
traveled, 𝑑, equals his initial speed times time, which is zero because he started
from rest, plus one half his acceleration times time squared.

His acceleration is the
acceleration we solved for earlier, 6.667 meters per second squared. And 𝑡 is 𝑡 one, 12.308
seconds. When we enter these values into
this equation and multiply them together, we find a distance of 504.9 meters. This is an important result; it
tells us that if we were to draw a sketch of Burt’s journey, then starting at the
beginning and for some portion of that journey, but not the whole thing, his
acceleration would be positive. Then for the remainder of the
journey, his speed would stay constant and his acceleration would be zero.

We can call the time during which
his acceleration is greater than zero 𝑡 one and the time for which his acceleration
is zero 𝑡 two. We don’t yet know whether 𝑡 two is
less than 𝑡 one, but we do know that if we add them together, 𝑡 one plus 𝑡 two
equals 𝑡, the total time we want to solve for. Since we’ve already solved for 𝑡
one, our task is to figure out 𝑡 two. We can begin by recalling that
average speed 𝑠 is equal to distance traveled divided by the time it takes to
travel that distance.

In our case, 𝑣 sub max, which is
equal to the remaining distance to travel which is the total course distance minus
the distance we solved for all divided by 𝑡 sub two. In the numerator 8.00 kilometers is
equal to 8000.0 meters. Looking at this equation, we see
that we can algebraically rearrange it by switching the position of 𝑣 sub max and
𝑡 sub two.

When we do that and then enter in
the value for 𝑣 sub max we solved for in meters per second, which is 82.05 meters
per second, then we can perform this division and solve for 𝑡 two, which is equal
to 91.35 seconds. So we now have values for 𝑡 one
and 𝑡 two. And when we add them together, we
get 𝑡, the time for the overall journey. 𝑡 one, 12.308 seconds, plus 𝑡
two, 91.35 seconds, gives us our total time for the journey 𝑡, which to three
significant figures is 104 seconds. This is the total time it took Burt
to travel the 8.00-kilometer course.