If a rocket with a mass of 3500 tons vertically accelerates at a rate of # 1/5 m/s^2#, how much power will the rocket have to exert to maintain its acceleration at 15 seconds?

Answer 1

#P_"thrust" = 5.26xx10^7# #"W"#

The given mass is assumed to be in metric tons.

We are asked to determine the power output that the rocket needs in order to maintain its acceleration for a predetermined period of time.

First, let's determine the rocket's mass in kilograms:

#m = 3500cancel("t")((10^3color(white)(l)"kg")/(1cancel("t"))) = ul(3.5xx10^6color(white)(l)"kg"#

Therefore, its weight in newtons is

#w = mg = (3.5xx10^6color(white)(l)"kg")(9.81color(white)(l)"m/s"^2) = ul(34335000color(white)(l)"N"#

We are informed that the rocket's acceleration is

#a = 0.2# #"m/s"^2#

Therefore, the net force acting on the rocket according to Newton's second law is

#sumF = ma = (3.5xx10^6color(white)(l)"kg")(0.2color(white)(l)"m/s"^2) = ul(7xx10^5color(white)(l)"N"#

There are just two forces influencing the rocket:

its mass (with a downward motion)

the thrust force acting upward that comes from the rocket

Using the net force equation, we can determine the thrust force:

#sumF = F_"thrust" - w#
#color(red)(F_"thrust") = sumF + w = 7xx10^5color(white)(l)"N" + 34335000color(white)(l)"N" = color(red)(ul(35035000color(white)(l)"N"#

We can now calculate the work that the thrust force does now that we know it:

#ul(W_"thrust" = F_"thrust" · s#
To find the displacement #s#, we can use a constant-acceleration equation:
#ul(s = v_0t + 1/2at^2#
The rocket started from rest, the initial velocity #v_0# is #0#, leaving us with
#ul(s = 1/2at^2#

We're given

#a = 0.2color(white)(l)"m/s"^2#
#t = 15color(white)(l)"s"#

Thus

#color(green)(s) = 1/2(0.2color(white)(l)"m/s"^2)(15color(white)(l)"s")^2 = color(green)(ul(22.5color(white)(l)"m"#

As a result, the thrust force's work is

#W_"thrust" = color(red)(35035000color(white)(l)"N") * color(green)(22.5color(white)(l)"m") = color(purple)(ul(7.883xx10^8color(white)(l)"J"#

The thrust's power output is provided by

#ul(P_"thrust" = (W_"thrust")/t#

Thus

#color(blue)(P_"thrust") = (color(purple)(7.883xx10^8color(white)(l)"J"))/(15color(white)(l)"s") = color(blue)(ulbar(|stackrel(" ")(" "5.26xx10^7color(white)(l)"W"" ")|)#
To maintain this acceleration for #15# #"s"#, the rocket's power output must be #color(blue)(5.26xx10^7color(white)(l)"watts"#.
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Answer 2

To calculate the power exerted by the rocket, you can use the formula:

Power = Force × Velocity

First, calculate the force exerted by the rocket using Newton's second law:

Force = Mass × Acceleration

Given: Mass = 3500 tons Acceleration = 1/5 m/s²

After finding the force, you need to calculate the velocity of the rocket at 15 seconds using the formula:

Velocity = Initial Velocity + (Acceleration × Time)

Given: Initial Velocity = 0 m/s (assuming the rocket starts from rest) Acceleration = 1/5 m/s² Time = 15 seconds

Once you have the force and velocity, you can find the power using the formula:

Power = Force × Velocity

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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