# Video: Calculating Instantaneous Acceleration Values of a Variable Mass Object

A spacecraft is moving uniformly in deep space where gravitational forces are negligible. At the instant 𝑡₀, the spacecraft’s engine is activated for a 30 s time interval, during which exhaust is ejected from the spacecraft at a rate of 2.0 × 10² kg/s with the exhaust moving at a speed of 2.5 × 10² m/s. The spacecraft’s mass before ejecting any exhaust is 2.0 × 10⁴ kg. 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 𝑡₀? What is the magnitude of the spacecraft’s acceleration at the instant 𝑡 = 15 s? What is the magnitude of the spacecraft’s acceleration at the instant 𝑡 = 30 s?

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### 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.

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