# Velocity versus Time

#### Velocity as function of time

You are given the dependence of velocity as function of time from the figure. Please find the position as function of time from the given graph, if the initial condition is $x(0)=0m$.  Also what is the position at 3s, 8s, and 10 s and the times when the slope of x(t) graph is zero.  Does it help to find the acceleration? Please specify the equation that you use for each segment:

1)  $Xf=Xi+Vix(t)+1/2ax(t^2)$

2)  $Vxf= Vxi + ax(t)$

3) $Vxf^2=Vxi^2+2a(Xf-Xi)$

Why do you use another equation for a different interval of time?

From $t=0$ to $t=3$ seconds using eq 2) for velocity $Vf =Vi+a*t$  it results an acceleration of $a=0 m/s^2$

Using eq 1) it results  $xf = xi+Vi*t = 0+4.5*3 = 13.5 m$ (the graph on this portion is a straight line with slope 4.5 m/s)

From $t=3$ to $t=8$ seconds using eq 2) for velocity $vf =vi+a*t$ we have

$-3 =4.5 +a*(8-3)$

$a =(-3-4.5)/5 = -1.5 m/s$

we can use eq 1) but modified with $t–> t-3$ in the form

$Xf = xi+vi*(t-3) +a*(t-3)^2/2$

which shows us that the grapf of $xi$ is a parabola. The parabola is down because the acceleration (coefficient of t^2 is negative) or we can use much simpler the equation

$vf^2 =vi^2 +2*a*(Xf-Xi)$

$Xf-Xi = (9-20.25)/(-1.5)/2 = 3.75 m$

$Xi =13.5 m$ found in the above step  and $Xf =3.75+13.5 =17.25 m$

From $t=8$ to $t=10$ seconds using eq 2) $vf=vi+a*t$ we have $a =0 m/s^2$

we use eq 1) modified $t –>(t-8)$

$xf =xi +vi*(t-8)$

$Xf=17.25 -3*(10-8) =17.25-6 =11.25 m$

It is useful to compute the acceleration to know which type of motion we have: uniform or uniform accelerated. For this is used eq 2) $vf=vi+a*t$
In general if the initial space xi is not zero it is easier to use the eq 3) $vf^2=vi^2+2*a*(xf-xi)$ instead of eq 1) because we need to alter the form of equation 1 by making a translation in time $t —>t-ti$