Motion Q (1 Viewer)

[shaon0]

A stone is let to fall from a building and 1 second later another stone is dropped vertically downwards with a velocity of 20m/s. When will the second stone overtake the first stone.

My working:
v_1=0+9.8t_1
v_2=20+9.8(t_2) where t_2=t_1 +1

 

[tommykins]

displacement of v1 = v2

solve for T, the answer you get is the time they meet each other.

so the time in which the second stoen overtakes the first stone is when t > T.

 

[vds700]

A stone is let to fall from a building and 1 second later another stone is dropped vertically downwards with a velocity of 20m/s. When will the second stone overtake the first stone.

My working:
v_1=0+9.8t_1
v_2=20+9.8(t_2) where t_2=t_1 +1

Integrate to find their displacementsm equate them and solve for t

 

[lolokay]

if acceleration were taking to be 10m/s (to make it easier), then after the first second, subbing into equations of motion:
dv/dt = 10
v = 10t +0
x = 5t² (assuming x(0) = 0)

v=10, x=5

at this point the second stone is 'dropped' with v=20, ie. velocity 10 higher than the first stone. Since the acceleration is constant for both, the velocity difference will always be 10m/s, so the second stone approaches the first with this velocity. Since the first stone only had a 5m head start, the second will catch it in 5/10 = 0.5 seconds.

 

[tommykins]

if acceleration were taking to be 10m/s (to make it easier), then after the first second, subbing into equations of motion:
dv/dt = 10
v = 10t +0
x = 5t² (assuming x(0) = 0)

v=10, x=5

at this point the second stone is 'dropped' with v=20, ie. velocity 10 higher than the first stone. Since the acceleration is constant for both, the velocity difference will always be 10m/s, so the second stone approaches the first with this velocity. Since the first stone only had a 5m head start, the second will catch it in 5/10 = 0.5 seconds.

Although you are correct, the question states -

A stone is let to fall from a building and 1 second later another stone is dropped vertically downwards with a velocity of 20m/s. When will the second stone overtake the first stone.

So wouldn't it be when t > T (where T is the time they meet each other) ?

 

[shaon0]

if acceleration were taking to be 10m/s (to make it easier), then after the first second, subbing into equations of motion:
dv/dt = 10
v = 10t +0
x = 5t² (assuming x(0) = 0)

v=10, x=5

at this point the second stone is 'dropped' with v=20, ie. velocity 10 higher than the first stone. Since the acceleration is constant for both, the velocity difference will always be 10m/s, so the second stone approaches the first with this velocity. Since the first stone only had a 5m head start, the second will catch it in 5/10 = 0.5 seconds.

We are meant to use v^2=u^2+2as, v=u+at or s=ut+0.5at^2

 

[shaon0]

Question:
An object is projected vertically upwards with a speed of 49m/s. Two seconds later another object is projected vertically upwards from the same spot at the same velocity. Find when and where the two objects will meet using either;
v^2=u^2+2as, v=u+at or s=ut+0.5at^2

 

[shaon0]

anyone?

 

[Darrow]

Those are the physics equations aren't they?
Why are you specified to use them?

 

[shaon0]

Those are the physics equations aren't they?
Why are you specified to use them?

These are mathematical equations which are used in physics. There interchangeable. Still it requires you to use them.

 

[lyounamu]

Question:
An object is projected vertically upwards with a speed of 49m/s. Two seconds later another object is projected vertically upwards from the same spot at the same velocity. Find when and where the two objects will meet using either;
v^2=u^2+2as, v=u+at or s=ut+0.5at^2

s1 = ut + 0.5at^2
= 49t + 0.5at^2

s2 = 49(t-2) + 0.5a(t-2)^2

49t + 0.5at^2 = 49(t-2) + 0.5a(t-2)^2
49t +0.5at^2 = 49t - 98 + 0.5at^2 - 2at + 2a
0 = -98 - 2at + 2a
-2at + 2a = 98
2a(1-t) = 98
a = -9.8
1-t = -5
t = 6

NOTE: I took acceleration as -9.8 since the question didn't specify whether it is -10, -9.81 or -9.8

By the way, in Mathematics Extension 1, there is no way that you will be asked this type of question. I really think Couchman fucked this topic up by introducing this part of projectile motion to the topic. It really isn't on the syllabus and it's from Physics aspect of Mathematics even though the formulas used are pretty similar though.

 

[shaon0]

s1 = ut + 0.5at^2
= 49t + 0.5at^2

s2 = 49(t-2) + 0.5a(t-2)^2

49t + 0.5at^2 = 49(t-2) + 0.5a(t-2)^2
49t +0.5at^2 = 49t - 98 + 0.5at^2 - 2at + 2a
0 = -98 - 2at + 2a
-2at + 2a = 98
2a(1-t) = 98
a = -9.8
1-t = -5
t = 6

NOTE: I took acceleration as -9.8 since the question didn't specify whether it is -10, -9.81 or -9.8

By the way, in Mathematics Extension 1, there is no way that you will be asked this type of question. I really think Couchman fucked this topic up by introducing this part of projectile motion to the topic. It really isn't on the syllabus and it's from Physics aspect of Mathematics even though the formulas used are pretty similar though.

Thanks...so should i be doing these questions or just skip them?

 

[lyounamu]

Thanks...so should i be doing these questions or just skip them?

To be honest, those questions are pretty good exercises to improve your knowledge on the difference between displacement, velocity, acceleration and some substitution skills. I actually didn't bother to finish them all but I did manage to finish half of that because I still wanted to do them as a light exercise.

BUT your main focus must be purely on the questions from projectile (especially on HOW TO DERIVE displacements and etc.)...but understand that it is just my opinion.

 

[shaon0]

To be honest, those questions are pretty good exercises to improve your knowledge on the difference between displacement, velocity, acceleration and some substitution skills. I actually didn't bother to finish them all but I did manage to finish half of that because I still wanted to do them as a light exercise.

BUT your main focus must be purely on the questions from projectile (especially on HOW TO DERIVE displacements and etc.)...but understand that it is just my opinion.

Alrite, i'll just finish about 0.75 of this exercise. thanks.