### Video Transcript

A spacecraft is moving uniformly in
deep space, where gravitational forces are negligible. At the instant 𝑡 sub zero, the
spacecraft’s engine is activated for a 30-second time interval, during which exhaust
is ejected from the spacecraft at a rate of 2.0 times 10 to the two kilograms per
second with the exhaust moving at a speed of 2.5 times 10 to the two meters per
second. The spacecraft’s mass before
ejecting any exhaust is 2.0 times 10 to the fourth kilograms. What magnitude force is applied to
the spacecraft by the engine while it is activated? What is the magnitude of the
spacecraft’s acceleration at the instant 𝑡 sub zero? What is the magnitude of the
spacecraft’s acceleration at the instant 𝑡 equals 15 seconds? What is the magnitude of the
spacecraft’s acceleration at the instant 𝑡 equals 30 seconds?

In this four-part exercise, we can
label the magnitude force applied by the spacecraft during activation, 𝐹 sub
𝑒. We can call the acceleration of the
spacecraft at time 𝑡 equals 𝑡 sub zero 𝑎 of 𝑡 sub zero. And we can make similar labels for
when 𝑡 equals 15 and 𝑡 equals 30 seconds.

In the problem statement, we’re
told the time rate of change in kilograms per second of fuel being burned by the
spaceship. We’re also told the speed of the
exhaust gases coming out of the engine and the initial mass of the spaceship. Because of the activation of the
engine creating exhaust gases and a fuel burn rate, we know that there’s a thrust
force created by the spaceship. This thrust force is equal to the
product of the exhaust gas speed times the time rate of change of mass.

Since we’re given both of these
values, we can plug in and solve for 𝐹 sub 𝑒. When we do and enter this product
on our calculator, we find, to two significant figures, it’s equal to 5.0 times 10
to the fourth newtons. That’s the thrust force that the
spacecraft produces.

Next, we wanna solve for the
acceleration of the spacecraft at various moments during its engine activation. We recall Newton’s second law of
motion that says that the net force acting on an object equals its mass times its
acceleration. When we write the forces acting on
our spaceship, we’re told that the force of gravity is negligible. So, it’s the thrust force, 𝐹 sub
𝑒, that equals the spacecraft’s mass times its acceleration. We’ll have to be careful with this
equation though because we know the mass of our spacecraft is not constant. It’s changing as fuel is burned
up. And therefore, its acceleration is
not constant either.

When 𝑡 is equal to 𝑡 sub zero,
the acceleration of our spacecraft is equal to the thrust force divided by the
initial mass, 𝑚 sub 𝑖, of the craft altogether. We solved for 𝐹 sub 𝑒 in part
one. And we’re given the mass of the
spacecraft initially. We’ve called it 𝑚 sub 𝑖. This fraction equals 2.5 meters per
second squared. That’s the spacecraft’s
acceleration at the outset of engine activation.

Next, we want to solve for the
acceleration of the spacecraft after its engines have been firing for 15
seconds. This is equal to the thrust force,
𝐹 sub 𝑒, divided by the mass of the spacecraft at this point in time. We’ve called it 𝑚 sub 15, where 𝑚
sub 15 is equal to the initial mass of the spacecraft minus 15 seconds times the
burn rate 𝑑𝑚 𝑑𝑡.

When we plug in for 𝑚 sub 𝑖 and
for 𝑑𝑚 𝑑𝑡, we find that the mass of the spacecraft after 15 seconds of fuel
burning is 17000 kilograms, down from its original 20000. When we plug in for 𝐹 sub 𝑒 and
𝑚 sub 15, calculating this fraction to two significant figures, we find a result of
2.9 meters per second squared. That’s the spacecraft’s
acceleration 15 seconds after turning on its engines.

Finally, we wanna solve for the
acceleration of the spacecraft 30 seconds after engine activation. This is equal to the thrust force,
𝐹 sub 𝑒, divided by the mass of the spacecraft at that point in time. 𝑚 sub 30 is equal to the initial
mass of the spacecraft minus 30 seconds times the fuel burn rate 𝑑𝑚 𝑑𝑡. This is equal to 14000
kilograms. Plugging in for 𝐹 sub 𝑒 and 𝑚
sub 30, we find an acceleration of 3.6 meters per second squared. That’s the spacecraft’s
acceleration after its engines have been running for 30 seconds.