Conversation started Jul 30, 2021 at 5:46.
Wolgwang
Wolgwang
Jul 30, 2021 05:46
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Magnitude of acceleration = Rate of change of speed
$\sqrt{a_x^2+a_y^2}=2\\a_y=0$
This is wrong?
Lllt
Lllt
Is the answer 0?
John Rennie
John Rennie
@Wolgwang That seems correct to me. I wonder if there is a mistake in the question as that seems too simple somehow.
Wolgwang
Wolgwang
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John Rennie
John Rennie
I'm going to make a coffee. I'll be a few minutes.
Lllt
Lllt
Jul 30, 2021 06:02
@Wolgwang wait I got it
Wolgwang
Wolgwang
I am listening :-)
Lllt
Lllt
the magnitude and direction of velocity both are changing at P
So there must be acceleration which is changing speed as well as direction
The direction is being changed by centripetal acceleration
the speed is being changed by tangential acceleration
satan 29
satan 29
wait i got it.
Wolgwang
Wolgwang
3 mins ago, by Wolgwang
I am listening :-)
satan 29
satan 29
$$\dfrac{d\sqrt{v_{x}^2+v_{y}^2}}{dt}=2$$
$$\dfrac{v_{x}a_{x}+v_{y}a_{y}}{\sqrt{v_{x}^2+v_{y}^2}}=2$$
where ax= d vx/dt and likewise for ay
$$\dfrac{6+4a_{y}}{5}=2$$
I think you can take it from here :)
Lllt
Lllt
Jul 30, 2021 06:14
Yes from this method we also get answer
Wolgwang
Wolgwang
@satan29 Thanks. I wonder why this is wrong...
Lllt
Lllt
@Wolgwang the rate of change of speed is not same as rate of change of velocity
satan 29
satan 29
@Wolgwang "acceleration is the rate of change of speed" is correct, and thats exactly what I used in my first line
Wolgwang
Wolgwang
brilliant.org/wiki/is-acceleration-the-rate-of-change-of-speed/…
satan 29
satan 29
hang on my bad
I messed up:
"magnitude of acceleration is not the rate of change of speed"
Lllt
Lllt
Jul 30, 2021 06:22
@Wolgwang have you studied circular motion?
satan 29
satan 29
the question mentions 2 as the rate of change of speed , and we are not supposed to interpret this as magnitude of acceleration. so sqrt(ax^2 +ay^2) is wrong
Prateek Mourya
Prateek Mourya
use the fact a.v/ magnitude v
satan 29
satan 29
$$ |\mathbf{a}|= |d/dt (\mathbf{v})|= |d/dt(v_{x} \hat{i} + v_{y} \hat{j})|= |a_{x}\hat{i}+a_{y}\hat{j}|= \sqrt{a_{x}^2+a_{y}^2} $$
Prateek Mourya
Prateek Mourya
= rate of change of speed
Wolgwang
Wolgwang
@Lllt A little...
@satan29 Hmm
satan 29
satan 29
Jul 30, 2021 06:26
however this is not the same as:
Prateek Mourya
Prateek Mourya
let acceleration be 2i+yj so rate of change of speed is the component of acceleratioon along velocity
satan 29
satan 29
$$d/dt (|\mathbf{v}|)= d/dt ( \sqrt{v_{x}^2 +v_{y}^2})= v_{x}a_{x} + v_{y}a_{y} / |\mathbf{v}|$$
In short:
$$| \mathbf{a} | \neq \dfrac{d | \mathbf{v}|}{dt}$$
Wolgwang
Wolgwang
:-o Thank you everyone
Lllt
Lllt
We can directly say that vector cannot be equal to a scalar @satan29
satan 29
satan 29
And also, as prateek said, the expression for d/dt (speed) can be simplified to:
$\mathbf{a}.\mathbf{v} / |\mathbf{v}|$
which makes a lot of intuitive sense too.
this is just the component of $\mathbf{a}$ in the direction of $\mathbf{v}$
The perpendicular component will not be responsible for changing the speed. It will change the direction though.
@Lllt well yes, but how is that relevant here?
@Wolgwang also, as Lllt mentioned, circular motion is a perfect example. In uniform circular motion, the speed is constant. however, there is acceleration of magnitude $v^2/R$
Lllt
Lllt
Jul 30, 2021 06:35
that acceleration(vector) will not be equal to rate of change of speed (scalar)
satan 29
satan 29
@Lllt We are talking about the magnitude of acceleration vector.
which is a scalar.
Lllt
Lllt
Ohk
satan 29
satan 29
thats just the expression we get when we differentiate $\sqrt{vx^2 + vy^2}$, are you fine with that?
Wolgwang
Wolgwang
NVM 😅
Yes, I got that. Thank you
 
Conversation ended Jul 30, 2021 at 6:40.