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.

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- Sep 5th 2006, 10:02 PMkolliBody
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 PMchanceyQuote:

Originally Posted by**kolli**

2. Does it change direction when it passes O?

3. What do you mean by initial displacement? - Sep 5th 2006, 11:32 PMCaptainBlackQuote:

Originally Posted by**kolli**

$\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 PMchanceyQuote:

Originally Posted by**CaptainBlack**

- Sep 6th 2006, 02:15 AMSoroban
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 AMCaptainBlackQuote:

Originally Posted by**chancey**

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

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 AMtopsquarkQuote:

Originally Posted by**CaptainBlack**

*describe*s0 in those terms: The initial displacement was 20 m to the left of the origin.

-Dan - Sep 6th 2006, 05:05 AMCaptainBlackQuote:

Originally Posted by**topsquark**

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 AMtopsquarkQuote:

Originally Posted by**CaptainBlack**

-Dan