How Do You Sketch the Correct Motion of a Parachute on a Velocity-Time Graph?

[Jimmy87]

Homework Statement

I have attached the homework question as it is on a graph so easier to see as attachment. I am stuck with last last part about sketching onto the graph the actual correct motion of the parachute

Homework Equations

None

The Attempt at a Solution

I have attached my attempt and the answer. My thinking is that the correct motion must start on the original line as this is acceleration due to gravity. It would then still accelerate but at a decreasing rate due to air resistance (which increase with speed) up to 12s. From 12s it would then decelerate in some time interval rather than go from 12s to 6s in zero time as the original graph shows. I can't tell how many marks I would get and don't really understand the answer. What does "decelerating phase parallel but sooner mean"? Parallel to what and sooner than what? And I have no idea what it means by "areas under graphs should be equal".

Thanks for any help offered.

 

[scottdave]

It looks like you are supposed to sketch what you think a real-life graph would look like, vs the model. Do you know what the area under the curve (velocity vs time) represents? Think about the units. Horizontal is seconds. Vertical is m/s. What happens when you multiply (m/s) by (s)? Can you relate this to some constraint in the original problem?
You are correct that the velocity would not change instantly. What would be the acceleration necessary for this to happen? What would that mean for the amount of force necessary?

 

[Jimmy87]

It looks like you are supposed to sketch what you think a real-life graph would look like, vs the model. Do you know what the area under the curve (velocity vs time) represents? Think about the units. Horizontal is seconds. Vertical is m/s. What happens when you multiply (m/s) by (s)? Can you relate this to some constraint in the original problem?
You are correct that the velocity would not change instantly. What would be the acceleration necessary for this to happen? What would that mean for the amount of force necessary?

Yes, the area gives the distance travelled. So would the first part (decreasing acceleration) have to reach a speed less than 12 m/s. If you draw this decreasing acceleration curve to 12s the area work out being much bigger otherwise?

I understand the second part - the vertical line means the time is zero which means the force and acceleration would be infinite. I'm just not sure how to accurately draw the deceleration when the parachute open. I know it asymptotically approaches 6 m/s but should it do this sooner or later? I still don't get what it means by "decelerating phase parallel but sooner"?

 

[scottdave]

It is going to be a steep change when the shoot opens (just not quite vertical), then change to a shallower pitch and approach the horizontal terminal velocity. Since the curve is above the horizontal line, then the first part (near freefall) will have to be below that line, for the total areas to be equal. It is supposed to be a sketch.

 

[Jimmy87]

It is going to be a steep change when the shoot opens (just not quite vertical), then change to a shallower pitch and approach the horizontal terminal velocity. Since the curve is above the horizontal line, then the first part (near freefall) will have to be below that line, for the total areas to be equal. It is supposed to be a sketch.

I think I get what you mean. So is this attached attempt better? Do the green and blue areas I have shaded need to be equal if I understand you correctly? They are not on the diagram but the only way to get them equal is if the decreasing acceleration (convex curve) barely curves at all.

 

[scottdave]

I think I get what you mean. So is this attached attempt better? Do the green and blue areas I have shaded need to be equal if I understand you correctly? They are not on the diagram but the only way to get them equal is if the decreasing acceleration (convex curve) barely curves at all.

That looks better, Except that the action (parachute open) should occur at the same time as the model, because I think the problem states that the parachute opens at a specific time. You should find that acceleration initially is slower due to air resistance (without the parachute there will be some). And yes, your blue area should be equal in area to the green (so that the total area under the curve equals 150 meters total distance).

 

[Jimmy87]

That looks better, Except that the action (parachute open) should occur at the same time as the model, because I think the problem states that the parachute opens at a specific time. You should find that acceleration initially is slower due to air resistance (without the parachute there will be some). And yes, your blue area should be equal in area to the green (so that the total area under the curve equals 150 meters total distance).

Thanks for your help. So more like this one i attached then? Could you just confirm the following reasoning is correct:

The maximum velocity in my sketch should be smaller than 12 m/s because if there is air resistance then after the same amount of time the velocity will be smaller so it would never get as high as 12 m/s in reality?

 

[scottdave]

That looks in about how I would sketch it.