Question

Closed

Answer 1

\[[AT ^{3} + Bt ^{2} +Ct +D\] is the equation

Answer 2

Which parameter--A, B, C or D--determines the initial velocity?

Answer 3

talk to me when you're back.

Answer 4

@JamesJ I didn't even get the question in this case,don't know what exactly is given :o

Answer 5

Suppose A = B = C = 0. What would be the initial velocity in this case?

Answer 6

0?

Answer 7

Yes, because the displacement is a constant, s(t) = D.

Now, suppose C is not zero and the displacement is given by
s(t) = Ct + D
What is the initial velocity now?

Answer 8

C?

Answer 9

Yes. For example, suppose
s(t) = 10t + 5 m, where t is in seconds.

Then the initial velocity is 10 m/s

Answer 10

yeah!

Answer 11

Now, suppose
s(t) = Bt^2 + D.

What now is the initial velocity?

Answer 12

B ?

Answer 13

No, zero. Here's a graph of
s(t) = t^2 + 1.
The velocity is the slope of the tangent to the graph. At t = 0, the slope is zero; the rate of change of displacement is zero.

Answer 14

But this article is clearly accelerating. The rate of acceleration for
s(t) = t^2 + 1
is
a = 2 m/s^2

Answer 15

Thus if the displacement is
s(t) = Bt^2 + 1, then the
initial velocity = 0 m/s
but
initial acceleration = 2B m/s^2

Answer 16

So now, returning to your general problem, if the displacement is given by
s(t) = At^3+ Bt^2 + Ct + D
what is the initial velocity and initial acceleration at t = 0?

Answer 17

Give me break :P

Answer 18

initial velocity = C
initial acc = 2B

Answer 19

at t =0

Answer 20

oh! so the answer is B and C, right?

Answer 21

Yes

Answer 22

@DLS that was quite straightforward. what did u not understand in this?

Answer 23

@JamesJ can u help us out in another prob? i'll tag u there.

Answer 24

I didnt get anything after this
Now, suppose
s(t) = Bt^2 + D.

What now is the initial velocity?

Answer 25

@DLS look at the sum from the beginning. u've been given an expression for displacement. u need initial velocity and accel. so first, differentiate the given expression to get exps for velocity and accel.

Answer 26

Here the initial velocity is zero. Look at the graph of the function I gave as an example,
s(t) = t^2 + 1.
The velocity is the slope of the tangent to that graph at t = 0. You can see the tangent at t = 0 is horizontal; i.e., the rate of change of displacement at exactly that point is .... zero, nothing. Hence the velocity at the moment is zero.

Answer 27

@DLS next, since we need 'initial' vel and accel, take t = 0. now take the ratio of whatever is left. u'll get the answer.

Answer 28

@DLS what JamesJ was doing is explaining to you 'what' the question is abt. since u mentioned that u didn't get the questn. so he was helping u out using various examples and cases.

Answer 29

okay..

Answer 30

so now, did u get 'what' the ques wants and how to get it?

Answer 31

To find velocity differentiate it once and to get acceleration differentiate it twice. To get the initial values just put t=0. You will get the answer.