Question

Another problem.

Answer 1

1 2 3?

Answer 2

1st

Answer 3

For 1 (a) use s=ut+1/2 at^2

Answer 4

where s=25m , u=0m/s , t=5s and you need to find a

Answer 5

I think the answer given for 1(a) is wrong..... unit of acceleration must be m/s^2
I got 2m/s^2

Answer 6

@sauravshakya , yeah, my answer is 2m/s^2 too.
I can't answer letter c and d of number 1.

Answer 7

For c you need to use v^2=u^2+2as

Answer 8

v = final velocity = 0m/s
s=18 m
and u =initial velocity

Answer 9

At first you need to find initial velocity (u) which is also the final velocity at the period when bus was accelerating

Answer 10

u=initial velocity = root(2*2*5) = root(20) m/s

Answer 11

Now, find a

Answer 12

v^2=u^2+2as
0^2 =(root20)^2 +2*a*18
a=-0.56 m/s^2

Answer 13

Is the given answer, incorrect? The answer should be -2.8m/s^2 right?

Answer 14

@sauravshakya , isnt u = 10 (velocity after accelerating for 5 sec at 2m/s^2)

Answer 15

Part B's answer's units is confusing, why is showing up in meters and not meters/sec when asking for velocity

Answer 16

oh yes
u=initial velocity = root(2*2*25)=10m/s

Answer 17

Thanx @dumbcow
I used time in place of distance

Answer 18

@OrionsBelt , yeah. But my answer there is 10m/s.

Answer 19

Now,
v^2 = u^2 +2as
0^2 = 10^2 +2*a*18

Answer 20

Is it 2.78?

Answer 21

@sauravshakya it is unclear. Why is is 2*a*18???

Answer 22

We don't know the time yet here.

Answer 23

v^2 =u^2 +2as
here s=distance

Answer 24

not time

Answer 25

\[v _{f}^{2} = v _{0}^{2} + 2a(\Delta x)\] from what I remember

Answer 26

It applies to constant acceleration, kinematics equations.

Answer 27

Ohhh. Sorry. I remembered already.

Answer 28

@sauravshakya , can you help me in problem 2?

Answer 29

@wio, any help?

Answer 30

Still need help? :)

Answer 31

Yes.

Answer 32

There's my cruddy drawing. Each "b" is the distance the elevator moved during the given time frame.

Answer 33

Yes. Following.

Answer 34

So, to start out, you can find how far the elevator went during t1 using the first equation of motion

b1 = vt + 1/2 a(t1)^2
b1 = 1/2a(t^2)

Answer 35

Which is where you can leave it for now. Next you want to find the velocity for the second part. How would you do that?

Answer 36

Like, what's the normal velocity equation given acceleration and time?

Answer 37

V=Vo+at?

Answer 38

yah. and Vo = 0, so we just have that our veloctiy

\[v=at_1\]

So you can use that in the second part to find b2

\[b_2 = v(t_2) = 4vt_1 = 4 (a t_1)t_1=4at_1^2\]

Answer 39

the only things that we're really "given" in the problem are height time, so we're trying to keep everything in terms of the thing we know and the thing we want, acceleration.

Answer 40

Is this making sense at all?

Answer 41

Yes. I know. Let me try \[b _{3} =at _{1}^{2}\]
Am I right?

Answer 42

fo sho

Answer 43

Then how would you find the final value of a?

Answer 44

Let me try again.
\[b _{1}+b _{2}+b _{3}=h\]
Therefore,
\[\frac{ 1 }{ 2 }at^2 +4at^2+at^2 = h\]
\[a (11t^2) = 2h\]
\[a =\frac{2h }{ 11t^2 }\], is my answer right? But the problem tells the answer should be a=h/5t^2

Answer 45

Oh! I'm sorry. I read your b3 wrong for some reason. How did you calculate it?

Answer 46

It should look quite similar to b1

Answer 47

I'm not so sure how I arrived there. Can you teach me? I know not the concept, honestly. Just let me understand the concept and I think I can beat those problems down. hahaha.

Answer 48

:D That's the right kind of fightin' spirit!

Answer 49

So we use the first equation of motion again,

\[b_3=vt_3 - \frac{1}{2}at_3^2\]

this time the acceleration the same magnitude, but is negative

Answer 50

We know to use that one because we know (well, kinda know. we know them in relation to each other) the velocity, acceleration and time.

Answer 51

unlike b1, however, our initial velocity isn't zero. It's the velocity it was traveling during part 2

Answer 52

Can you just continue it? And then I'll try to understand.

Answer 53

from part 2 we know

\[vt_3 = vt_1 = at_1^2\]

so then we put that into the equation of motion

\[b_3=at_1^2 - \frac{1}{2}at_1^2 = \frac{1}{2}at_1^2\]

Answer 54

Then what you did before was correct. b3+b2+b1 = h

Answer 55

Great! Thank you so much. I already understood this problem. :))

Answer 56

@AllTehMaffs Just a question, earlier you stated \[b_3 =vt_3-\frac{ 1 }{ 2 }a _{2}^{3}\] and then you had this \[b_3 = a t_{2}^{1}-\frac{ 1 }{ 2 }a _{1}^{2}\] I understand the swapping of the 'vt_3' to 'at_2^1' but how did you get the 2nd part?

Answer 57

Oh nvm. got it. Thanks !

Answer 58

^^