googly from ATGS 20

for beginners or class 11 people...

easy for ones knowing calculus...

a particle is moving with constant speed in the x-y plane...

the trajectory is given by the path is

y=ax3+bx2+cx+d

find the point of time when acceleration is zero...

*EDITS:

1) yeah... take the constant speed to be v!!

2) take the motion starts from the origin...

11 Answers

11
Khyati ·

How ppl of class 11 know calculas? :P

11
Khyati ·

well it would be those points for which the double differentiation with respect to t will be

zero, i.e, d2x/dt2 = 0

So the answer to the question

the trajectory is given by the path is

y=ax3+bx2+cx+d

find the point of time when acceleration is zero... is

the point c and d,

Hope am correct :)

71
Vivek @ Born this Way ·

I'm really Confused.

1. Y(x) is given. How are they related to time.
2. @ x=0 , y=d. So how does the particle starts from Origin (0,0)

well it would be those points for which the double differentiation with respect to t will be

zero, i.e, d2x/dt2 = 0

Correct, But Elaborate the Process how you reached there. I don't agree with your answer though!

49
Subhomoy Bakshi ·

@khyati...i do not think that u are quite right!! :(

11
Khyati ·

Ohh!! I am sorry I made a mistake, but now I don't know how to go about it. :(

Subhomoy you help, Please!

Would it be d2y/dt2 = 0, I am too confused now.

11
Khyati ·

am attempting once more,

dy/dt =d(y=ax3+bx2+cx+d)/dt

y' =3ax2+2bx + c + 0 (since differentiation of the constant term is zero)

Now, differentiating y' w.r.t t

dy'/dt =d(3ax2+2bx + c)/dt

y" =6ax + 2b

Therefore, d2y/dt2 =6ax + 2b

So the points which have acceleration zero are c and d

11
Khyati ·

Differentiation of

xn = nxn-1, I made mistake in this differentiation I guess

49
Subhomoy Bakshi ·

well....the equation given gives the trajectory of the particle....

speed is constant throughout....

thus velocity will be constant (i.e. acceleration is zero) when the path is straight

i.e. slope is constant!!!

11
Khyati ·

Ohhhhhh!!!!!! I just didn't read that. I only saw the equation and started making mistakes.

Thanks Subho :)

From next time I'll read the whole question and then attempt it :)

49
Subhomoy Bakshi ·

:P

try again....

71
Vivek @ Born this Way ·

dy'/dt =d(3ax2+2bx + c)/dt

Well You cannot differentiate x w.r.t t. Because 'x' is abscissa and NOT the displacement that you can differentiate w.r.t t.

Subho, I thought exactly that BUT I thought you're never wrong. So I started thinking in other way. Really, I need to build up confidence because I solve 90% problems correct but don't know whether I have solved them correctly?!!

Feeling Better as for Now!

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