1. ## 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!!

2. 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?

3. 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

4. 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?

5. Hello, kolli!

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

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" $\displaystyle O.$

6. Originally Posted by chancey
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

7. 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

8. 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

RonL

9. 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

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