Solving Velocity Problems on the Moon

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In summary, the conversation is about a velocity problem involving an arrow shot upward on the moon. The formula for its height in meters after t seconds is H=58t-.83t^{2}. The questions asked include finding the velocity of the arrow after one second, when t=a, and when it will hit the moon. The solution involves setting H=0 and solving for t to find the corresponding velocity, or using the principle of energy conservation to determine that the speed remains the same but the arrow will be headed downward when it hits the moon.
  • #1
member 5645
Calc help - velocity problem...

Just looking to be pointed in the right direction, for now. I understand the equations I have learned about velocity, but am having trouble setting up this word problem:

If an arrow is shot upward on the moon, with a velocity of 58m/s, it's heigt in meters after t seconds is given by
[tex]H=58t-.83t^{2} [/tex]

Find the velocity of the arrow after one second? (I already figured this out)
Find the velocity of the arrow when t=a?
When will the arrow hit the moon? (I already figured this out)
With what velocity will the arrow hit the moon?
 
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  • #2
Set H = 0, solve for t then find the corresponding velocity.

Or, since energy is conserved and there is no air resistance then the kinetic energy will be the same at the end of the trajectory as it was at the beginning of the trajectory since the potential energy has a single value at the surface - therefore the speed is the same but the arrow is headed downward.
 
  • #3


To find the velocity of the arrow when t=a, we can simply plug in a for t in the equation H=58t-.83t^{2}. So, the velocity at t=a would be 58a-.83a^{2}.

To find when the arrow will hit the moon, we can set the equation equal to 0 and solve for t. So, 58t-.83t^{2}=0. This gives us two solutions, t=0 and t=69.88 seconds. Since t=0 represents the initial position, we can disregard it and say that the arrow will hit the moon after approximately 69.88 seconds.

To find the velocity at which the arrow hits the moon, we can plug t=69.88 into the equation for velocity, which would give us a velocity of 58(69.88)-.83(69.88)^{2}=1957.04 m/s.

I hope this helps point you in the right direction and gives you a better understanding of how to approach velocity problems. Remember to always carefully read the problem and identify what information you have and what you need to find. Then, use the appropriate equations and plug in the given values to solve for the unknown variable. Good luck with your calculations!
 

FAQ: Solving Velocity Problems on the Moon

1. How do I calculate velocity?

Velocity is calculated by dividing the change in position by the change in time. In other words, it is the rate at which an object changes its position.

2. What are the units for velocity?

The most common units for velocity are meters per second (m/s) or kilometers per hour (km/h). Other units used for velocity include feet per second (ft/s) and miles per hour (mph).

3. What is the difference between velocity and speed?

Velocity is a vector quantity that includes the direction of motion, while speed is a scalar quantity that only describes how fast an object is moving. In other words, velocity takes into account the direction of motion while speed does not.

4. How do I calculate average velocity?

Average velocity is calculated by dividing the total displacement of an object by the total time it takes to travel that distance. This can be represented by the equation v=Δx/Δt, where v is average velocity, Δx is change in position, and Δt is change in time.

5. Can velocity be negative?

Yes, velocity can be negative. This indicates that the object is moving in the opposite direction of the chosen positive direction. For example, if a car is moving west and the positive direction is east, then the velocity of the car would be negative.

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