Kinematics of Linear Motion: Solving Problems with Constant Acceleration

[freshbox]

Homework Statement

I am unable to solve part b and please help me check whether is there anything wrong with my graph..

The Attempt at a Solution

Segment A
Vi=0
V=3
S=60
a=9.81

v=vi+at
3=0+9.8(t)
t=0.305s

Segment B
Vi=3
V=0
a=-2

v=vi+at
0=3-2t
t=1

t=1+0.305=wrong answer

 

[Astronuc]

One's velocities are incorrect.

The parachutist begins free falling with a velocity v=0 and falls a distance of 60 m. One must determined the velocity (or rather speed) after free falling 60 m.

The speed of 3 m/s is the speed at which the parachutist lands.

Please refer to this - http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#ffall

and http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html#mot1

 

[azizlwl]

I think you should draw the diagram roughly to scale.
The free fall acceleration should be 9.8m/s² Thus in 1 sec. the velocity should be changing from 0 to 9.8m/s.

As you see from the graph the area first triangle is equal to 60m.
From simple formula of triangle you find the height(the velocity)

From your diagram the time= 40 sec(60=1/2vt). which wrong. After 40 sec. the velocity will be more than 3m/s.

 

[freshbox]

azizlwl: So the tip of my triangle should have abit of curve?

Astronuc: Which of my velocity is incorrect? The 2 website that you listed is for me to check my answer?

 

[azizlwl]

First try to find all the variables in segment A.
Acceleration=?
Displacement=?
Initial velocity=?
final velocity =?
Time taken=?

What equation should you use to find the final velocity if you have the values of acceleration and displacement?

 

[freshbox]

Segment A
Vi=0
V=3
S=60
a=9.81

v=vi+at
3=0+9.8(t)
t=0.305s

What equation should you use to find the final velocity if you have the values of acceleration and displacement?

I would use v^2 = vi^2 +2as

 

[Doc Al]

Segment A
Vi=0
V=3
S=60
a=9.81

v=vi+at
3=0+9.8(t)
t=0.305s

I assume segment A is the free fall segment. The 3 m/s is the final speed at the end of segment B--it has nothing (directly) to do with segment A.

What equation should you use to find the final velocity if you have the values of acceleration and displacement?

I would use v^2 = vi^2 +2as

That's the equation you need to figure out the final speed at the end of segment A.

 

[freshbox]

Hey Doc Al, thanks for answering :)

"The parachute opens he experiences a deceleration of 2m/s^2 until he reaches the ground with a velocity of 3m/s"

Shouldn't it be 3m/s at the start of segment B and Final V = 0 since all things that reach the ground should have 0 speed because it has landed

So if you are saying "The 3 m/s is the final speed at the end of segment B" that means my velocity time graph is wrong already.

 

[Doc Al]

"The parachute opens he experiences a deceleration of 2m/s^2 until he reaches the ground with a velocity of 3m/s"

Shouldn't it be 3m/s at the start of segment B and Final V = 0 since all things that reach the ground should have 0 speed because it has landed

No. When they say that it reaches the ground with some speed, they mean speed just before it hits. (Once it hits the ground other forces come into play, but we don't care about that motion.)

So if you are saying "The 3 m/s is the final speed at the end of segment B" that means my velocity time graph is wrong already.

That's exactly what I'm saying.

The first step is to correctly find the speed at the end of segment A (which will also be the speed at the beginning of segment B).

 

[freshbox]

Sorry i want to ask a stupid question. How do you see if the question has segments in it?

 

[freshbox]

Doc Al, can you help me take a look at my velocity time graph.. I think it's wrong because it looks abit weird to me The circle part i don't know where to end it since the question says reaches the ground with a 3/ms So it cannot be 0m/s. And the initial velocity is 0 so my point has to stopped there. Is it right?


PS: Btw i managed to solved this question thanks to you...

 

[Doc Al]

Sorry i want to ask a stupid question. How do you see if the question has segments in it?

One way is to ask yourself if you can describe the motion with a single acceleration (and thus a single equation). Here you are told that the acceleration changes, so it makes sense to break the motion into constant acceleration segments.

 

[Doc Al]

Doc Al, can you help me take a look at my velocity time graph.. I think it's wrong because it looks abit weird to me The circle part i don't know where to end it since the question says reaches the ground with a 3/ms So it cannot be 0m/s. And the initial velocity is 0 so my point has to stopped there. Is it right?

Looks fine to me except that you can make it a bit closer to scale. The final speed is 3 m/s, but compare its height above the zero mark with the 34.31 m/s height. Other than that, it's good.

PS: Btw i managed to solved this question thanks to you...

Yay! Good work.

 

[freshbox]

Thanks, thanks a lot for taking your time to help me.
Thanks to azizlwl & Astronuc too!:tongue: