# Body

• Sep 5th 2006, 10:02 PM
kolli
Body
A body moves in a straight line with an acceleration 10m/s^2. If after 2 s it passes through O and after 3s it is 25 m from O, find its initial displacement relative to O.

Help thanks!! :)
• Sep 5th 2006, 10:48 PM
chancey
Quote:

Originally Posted by kolli
A body moves in a straight line with an acceleration 10m/s^2. If after 2 s it passes through O and after 3s it is 25 m from O, find its initial displacement relative to O.

Help thanks!! :)

1. Does it travel for 3 seconds after is passes O, or from the start?

2. Does it change direction when it passes O?

3. What do you mean by initial displacement?
• Sep 5th 2006, 11:32 PM
CaptainBlack
Quote:

Originally Posted by kolli
A body moves in a straight line with an acceleration 10m/s^2. If after 2 s it passes through O and after 3s it is 25 m from O, find its initial displacement relative to O.

Help thanks!! :)

The displacement as a function of time is:

$\displaystyle s=5t^2+v_0t+s_0$

where $\displaystyle v_0$ is the initial velocity, and $\displaystyle s_0$ is the initial displacement.

RonL
• Sep 5th 2006, 11:56 PM
chancey
Quote:

Originally Posted by CaptainBlack
The displacement as a function of time is:

$\displaystyle s=5t^2+v_0t+s_0$

where $\displaystyle v_0$ is the initial velocity, and $\displaystyle s_0$ is the initial displacement.

RonL

Can you show us how to use the formula?
• Sep 6th 2006, 02:15 AM
Soroban
Hello, kolli!

I'll assume this is a Calculus problem and you're familiar with integration.

Quote:

A body moves in a straight line with an acceleration $\displaystyle 10\,\text{m/s}^2$
After 2s it passes through $\displaystyle O$, and after 3s it is 25 m from $\displaystyle O$.
Find its initial displacement relative to $\displaystyle O.$

I further assume that $\displaystyle O$ is the origin: $\displaystyle x = 0$

Let $\displaystyle x(t)$ be the position function for the body.

Acceleration is the derivative of the velocity: .$\displaystyle a(t) \:=\:v'(t)\:=\:10$
. . Integrate: .$\displaystyle v(t)\:=\:10t + C_1$

Velocity is the derivative of the position: .$\displaystyle v(t) \:= \:x'(t)\:=\:5t^2 + C_1$
. . Integrate: .$\displaystyle x(t)\:=\:5t^2 + C_1t + C_2$

We are told that: $\displaystyle x(2) = 0$
. . $\displaystyle 5\cdot2^2 + C_1\cdot2 + C_2\:=\:0\quad\Rightarrow\quad 2C_1 + C_2\:=\:-20$ . [1]

We are told that: $\displaystyle x(3) = 25$
. . $\displaystyle 5\cdot3^2 + C_1\cdot3 + C_2\:=\:25\quad\Rightarrow\quad 3C_1 + C_2 \:=\:-20$ . [2]

Subtract [1] from [2]: .$\displaystyle C_1 = 0$

Substitute into [1]: .$\displaystyle 2\cdot0 + C_2 \:=\:-20\quad\Rightarrow\quad C_2 = -20$

Hence, the position function is: .$\displaystyle x(t)\:=\:5t^2 - 20$

Therefore, at $\displaystyle t = 0,\;x(0) = -20$
. . Initially, the body was 20 meters "behind" $\displaystyle O.$

• Sep 6th 2006, 02:18 AM
CaptainBlack
Quote:

Originally Posted by chancey
Quote:

Originally Posted by CaptainBlack
The displacement as a function of time is:

$\displaystyle s=5t^2+v_0t+s_0$

where $\displaystyle v_0$ is the initial velocity, and $\displaystyle s_0$ is the initial displacement.

RonL

Can you show us how to use the formula?

You are told that when $\displaystyle t=2$ $\displaystyle s=0$ so:

$\displaystyle 0=5\times 2^2+v_0 \times 2+s_0$,

and when $\displaystyle t=3$ $\displaystyle s=25$, so:

$\displaystyle 25=5\times 3^2+v_0 \times 3+s_0$.

These constitute a pair of simultaneous equations for $\displaystyle v_0$ and $\displaystyle s_0$.

Solve these and you will have found the initial displacement $\displaystyle s_0$ (which
will be negative with the sign convention used here, change its sign if
you think that the questioner wanted it positive).

RonL
• Sep 6th 2006, 04:54 AM
topsquark
Quote:

Originally Posted by CaptainBlack
You are told that when $\displaystyle t=2$ $\displaystyle s=0$ so:

$\displaystyle 0=5\times 2^2+v_0 \times 2+s_0$,

and when $\displaystyle t=3$ $\displaystyle s=25$, so:

$\displaystyle 25=5\times 3^2+v_0 \times 3+s_0$.

These constitute a pair of simultaneous equations for $\displaystyle v_0$ and $\displaystyle s_0$.

Solve these and you will have found the initial displacement $\displaystyle s_0$ (which
will be negative with the sign convention used here, change its sign if
you think that the questioner wanted it positive).

RonL

Since the original question was asking for a displacement, which is a vector, probably the best thing to do would be to define a positive direction (say, to the right), and describe s0 in those terms: The initial displacement was 20 m to the left of the origin.

-Dan
• Sep 6th 2006, 05:05 AM
CaptainBlack
Quote:

Originally Posted by topsquark
Since the original question was asking for a displacement, which is a vector, probably the best thing to do would be to define a positive direction (say, to the right), and describe s0 in those terms: The initial displacement was 20 m to the left of the origin.

-Dan

The sign convention (sense of the vectors) is implicit defined by taking the
acceleration to be +ve. Whether this is to the right or left is in a sense
irrelevant to us, it is in the direction of (if +ve) the acceleration or in the
opposite (if -ve) direction to the acceleration.

Maybe this is too sophisticated an idea for our posters, if so I will concede the
point :cool:

RonL
• Sep 6th 2006, 05:10 AM
topsquark
Quote:

Originally Posted by CaptainBlack
The sign convention (sense of the vectors) is implicit defined by taking the
acceleration to be +ve. Whether this is to the right or left is in a sense
irrelevant to us, it is in the direction of (if +ve) the acceleration or in the
opposite (if -ve) direction to the acceleration.

Maybe this is too sophisticated an idea for our posters, if so I will concede the
point :cool:

RonL

Actually, no, I agree with you completely. Typically in any Physics I class students are taught to use what I call the "1-D vector notation" to describe vectors. I just wanted to point out the vector nature of the answer to highlight that the displacement is, in fact, a vector which students tend to forget about rather quickly. My last post was an example of me being (slightly) anal about the solution statement. :)

-Dan