A bullet in a gun is accelerated
from the firing chamber to the end of the barrel at an average rate of 6.20 times 10
to the fifth meters per second squared for a time of 8.10 times 10 to the negative
four seconds. What is the speed of the bullet
when it leaves the gun’s barrel?
The first thing we can do here is
draw a picture of what we have going on. So we have the barrel of our gun
and the firing chamber, and we’re told that our bullet starts out here in the firing
chamber and then over time makes its way down the barrel, and eventually the ball
leaves the barrel with some final velocity that will call 𝑣 sub 𝑓. What we want to solve for is that
final velocity of the bullet as it leaves the gun’s barrel.
We’re given some information about
how the bullet is accelerated over this time. We’re told that as the bullet is in
the barrel, it experiences an average acceleration — we’ll call in 𝑎 sub avg for
average acceleration — of 6.20 times 10 to the fifth meters per second squared.
And we’re also told that the time,
𝑡, that the bullet is being accelerated in the barrel is equal to 8.10 times 10 to
the negative fourth seconds. To approach this problem, let’s
recall a definition; it relates acceleration to velocity, and what it says is that
acceleration is defined as a change — we’ll use the Greek symbol Δ to represent
change — a change in velocity divided by the change in time.
We can apply this definition to our
own scenario, where 𝑎 will be given as 𝑎 sub avg. We have an initial velocity of the
bullet we’ll call 𝑣 sub 𝑖. We have a final velocity of the
bullet we wanna solve for. And if we take that difference,
that’s the change in velocity. And all that is divided by what
we’ll call 𝑡, which is the time that the bullet is accelerating in the barrel.
Now if we take a look at this
equation in blue you can see that 𝑣 sub 𝑓 is what we’re trying to solve for; 𝑡
we’re given, but what about 𝑣 sub 𝑖? What is that? Well we can assume that the bullet
starts from rest while it’s in the chamber, that at that point its velocity is zero,
meaning that our initial velocity, 𝑣 sub 𝑖, is zero meters per second. What this means is we now know all
the elements in this equation.
We want to rearrange this equation
so that it says 𝑣 sub 𝑓 equals something. We’re not sure what yet, but we’ll
figure that out. But before we do, let’s start off
by removing 𝑣 sub 𝑖 because we already know that that is zero. That gives us an equation which
says 𝑎 sub avg is equal to 𝑣 sub 𝑓, the final velocity the bullet as it leaves
the barrel, divided by 𝑡.
Now again, we would like an
equation that says 𝑣 sub 𝑓 equals something else. So to get there, let’s multiply
both sides of the equation we have by 𝑡. And we’ll see that on the
right-hand side, the 𝑡 in the numerator and denominator cancel out and we’re left
with an equation that does indeed give us what we want 𝑣 sub 𝑓 is equal to 𝑎 sub
avg, the average acceleration, multiplied by 𝑡.
From here, it’s a matter of
plugging into this equation the variables we’ve been given. We know 𝑎 sub avg and we know 𝑡
because they’re given to us in the problem statement. So when we plug those values in, 𝑎
sub avg equals 6.20 times 10 to the fifth meters per second squared and 𝑡 is equal
to 8.10 times 10 to the negative fourth seconds.
Now before we multiply these
numbers together, notice something about the number of significant figures we’re
working with. In our first value, we have three
significant figures; and in our second value, we also have three significant
figures. That means our final answer will
also have three significant figures.
When we multiply 6.20 times 10 to
the fifth meters per second squared times 8.10 times 10 to the negative fourth
seconds, we end up with the final velocity of 502 meters per second.
That is equal to the final velocity
of the bullet as it leaves the barrel of the gun.