I have tried this problem over 4 times and keep getting the exact same answer, which is incorrect. Could someone please help by pointing out what I did wrong, or letting me know if I am actually on the completely wrong path?
Thanks!
Have a wonderful day :)
Hi! everyone! ;)
I have a problem with the development of this problem.
I need to resolve it with 2 procedures: trigonometry and instant centers. My advance can be see in the next image:
The instant centers procediment its (1) up and trigonometry procediment its (2) down.
I know that the...
i have tried V=Vi - (g*t) with an answer equals to 4
as well as
d=Vi(t) + 1/2 a t^2 with an answer equals to 19
my final asnwer was 1.43 and it was still wrong
i used formula a=v/t
v=14
t=10
i divided 14/10 to get 1.4
Here's what I did so far.
The velocity of the first car is ##v = v_0 +at##
Frame of reference S = the road
Frame of reference S' = the second car
thus, v' is the speed of the first car in the frame of reference S' and v the speed in the frame of reference S.
Here's what make me doubt.
The...
Was wondering if acceleration seems to be a more fundamental property/quantity in the universe as compared to velocity or distance because acceleration can be defined in more absolute terms in a frame depending on the forces acting inside that reference frame.
Considering a very simple example...
According to this link here https://en.wikipedia.org/wiki/Relativistic_mechanics#Force , we can inverse the relation of force in terms of velocity and acceleration:
$$
\mathbf{F} = \frac{m\gamma^3}{c^2}(\mathbf{v} \cdot \mathbf{a})\mathbf{v} + m\gamma\mathbf{a}
$$
to get:
$$
\mathbf{a} =...
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>
>A stuntman jumped from $1.25 \ \text{m}$ height and, landed at distance $10 \ \text{m}$. Find velocity when he jumped. (Take $\text{g}=10 \ ms^{-2}$)
I had solved it following way.
$$h=\frac{1}{2}gt^2$$
$$=>1.25=5\cdot t^2$$
$$=>t=\frac{1}{2}$$
And, $$s=vt$$
$$v=\frac{s}{t}$$
$$=\frac{10 \...
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$\tiny{1.2.1}$
An object is propelled vertically upward with an initial velocity of 20 meters per second.
The distance s (in meters) of the object from the ground after t seconds is
$s=-4.9t^2+20t$
(a) When will the object be 15 meters above the ground?
$15=-4.9t^2+20 \implies -4.9t^2 =-5$
ok...
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And then after that the velocity will be constant (s = v*t)
45/t2
Now i am stuck here because i have 3 equations but 4...
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b) What are their velocities if the...