# Video: Applying Kinematic Formula to a Body Subject to a Constant Net Force Directed against Gravity

A large rocket has a mass of 2.00 × 10⁶ kg at takeoff, and its engines produce a thrust of 3.50 × 10⁷ N. Find the rocket’s initial vertically upward acceleration. How much time is required for the rocket to attain a vertically upward velocity of 120 km/h, assuming constant mass and thrust?

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### Video Transcript

A large rocket has a mass of 2.00 times 10 to the sixth kilograms at takeoff, and its engines produce a thrust of 3.50 times 10 to the seventh newtons. Find the rocket’s initial vertically upward acceleration. How much time is required for the rocket to attain a vertically upward velocity of 120 kilometers per hour, assuming constant mass and thrust?

In this problem, we’ll not only assume as the problem states that the mass and thrust of the rocket are constant; we’ll also assume that 𝑔, the acceleration due to gravity, is exactly 9.8 meters per second squared.

In this problem statement, we’re told that the rocket has a mass of 2.00 times 10 to the sixth kilograms and that its engines produce a thrust of 3.50 times 10 to the seventh newtons. The rocket begins at rest but eventually attains a vertically upward velocity of 120 kilometers per hour.

The problem has two parts. In the first part, we wanna find the rocket’s initial vertically upward acceleration; we’ll call that 𝑎 sub 𝑣. In part two, we’re asked to solve for how much time is required to move the rocket from an initial speed of zero to a vertically upward velocity of 120 kilometers per hour; we’ll call that variable 𝑡.

To begin solving this problem, let’s draw a diagram of what’s going on. We have in this situation a rocket taking off from an initial velocity of zero. The rocket has a mass of 2.00 times 10 to the sixth kilograms.

We’re told that the thrust created by the rocket’s engines — we’ll call that 𝐹 sub 𝑡 for the thrust force — is equal to 3.50 times 10 to the seventh newtons.

We want to solve for the acceleration of the rocket. To do that, let’s recall Newton’s second law of motion. The second law states that the net force on an object equals the mass of that object times its acceleration. Let’s apply this equation to our scenario.

To do that, let’s first draw on this diagram of the rocket all the forces that are at play. Recall that this is called a free body diagram, where we draw in vectors to represent every force that is acting on the rocket.

One force that acts on the rocket and all masses is the force of gravity; we’ll call that 𝐹 sub 𝑔. Another force that acts on the rocket is the thrust force, 𝐹 sub 𝑡. We know because the rocket is accelerating upward and leaving the ground that the thrust force exceeds the force of gravity, so we draw that vector longer than the vector for gravity.

Let’s choose to make the positive direction in our diagram up, meaning that motion up is positive motion and motion down we’ll call negative. So 𝐹 sub 𝑡 acts in the positive direction and 𝐹 sub 𝑔 in the negative.

If we now write out a Newton’s second law expression, as it relates to our rocket leaving the ground, it would look like this: 𝐹 sub 𝑡 minus 𝐹 sub 𝑔 equals 𝑚𝑎, where 𝑚 is the mass of our rocket and 𝑎 is its vertical acceleration.

Now looking at this equation, we’ve been given 𝐹 sub 𝑡, the thrust force; what’s 𝐹 sub 𝑔? Recall that the weight force of an object, represented by a capital 𝑊, is equal to the mass of that object times 𝑔, the acceleration due to gravity. So we can replace 𝐹 sub 𝑔 in this equation with 𝑚𝑔, where 𝑚 is the mass of the rocket.

We want to isolate 𝑎 sub 𝑣, the vertical acceleration of the rocket. To do that, let’s divide both sides of the equation by 𝑚. The mass term on the right-hand side of our equation then cancels out, leaving us with 𝑎 sub 𝑣 by itself. 𝑎 sub 𝑣 is equal to the thrust force minus the weight force divided by the mass of the rocket; we can now plug in for each one these values.

𝐹 sub 𝑡, the thrust force of the rocket, is 3.50 times 10 to the seventh newtons and the rocket’s mass is 2.00 times 10 to the sixth kilograms and 𝑔 we treat as exactly 9.8 meters per second squared.

When we enter these numbers into our calculator, we find an acceleration of 7.70 meters per second squared. That’s the net acceleration of the rocket vertically upward under the influence of the thrust force as well as gravity.

Now we move to part two, where we seek to solve for the time it will take for the rocket to accelerate from moving at zero meters per second to 120 kilometers an hour.

As a first step to figuring this out, let’s convert the final speed of the rocket, currently in kilometers per hour, to units of meters per second, which will work nicely with other calculations. So 𝑣 sub 𝑓, the final speed of the rocket, is equal to 120 kilometers an hour. We want to convert this to units of meters per second. To do that, we want to change the distance units from kilometers to meters and the time units from hours to seconds.

Let’s do the distance units first. There are 1000 meters in one kilometer. When we multiply this conversion factor through, we see the units of kilometers cancel. And we’re left with distance units of meters, just as we want.

Now on to units of time. We know that there are 3600 seconds in one hour. So if we multiply this term by the fraction, one hour divided by 3600 seconds, then when we look to see how our units multiply out, we see the units of hours cancel, and the units of seconds remain.

This means that if we multiply through from left to right with all our conversion factors, we’ll wind up with a final velocity, now stated in units of meters per second. And when we do that calculation, we find a speed of 33.33 meters per second.

Now we have the final velocity of our rocket in meters per second and we also know the initial velocity in meters per second, which is zero. We want to solve for the time it takes the rocket to move from this initial to the final velocity under an acceleration of 7.70 meters per second squared.

Since the acceleration is constant, the kinematic equations we’ve seen before apply to this scenario. Let’s recall those equations. These are the four kinematic equations which all apply when acceleration is constant, like in our scenario.

We want to find an equation that will help us solve for 𝑡, given that we know 𝑎, the final velocity, and the initial velocity. The first equation listed here, 𝑣 sub 𝑓 equals 𝑣 sub 𝑖 plus 𝑎𝑡, is a good fit.

When we apply it to our situation, we see that 𝑣 sub 𝑓 is what we solved for, 33.33 meters per second; 𝑣 sub 𝑖 is zero; 𝑎 sub 𝑣 is the vertical acceleration that we solved for in part one; and 𝑡 is the variable we’re solving for now.

We can divide both sides of this equation by 𝑎 sub 𝑣, which cancels that term out of the right side of our equation. We can now rewrite this equation to be expressed as 𝑡 equals 𝑣 sub 𝑓 divided by 𝑎 sub 𝑣, and we can now plug in for 𝑣 sub 𝑓 33.33 meters per second and for 𝑎 sub 𝑣 7.70 meters per second squared.

When we calculate this fraction, we find a time value of 4.33 seconds. That’s the time the rocket needs to accelerate from rest to a final speed of 120 kilometers per hour.