# Homework Help: Kinematics, getting position and velocity equations from an acceleration graph

1. Aug 23, 2010

### Telemachus

1. The problem statement, all variables and given/known data
I have some doubts to the next exercise. It reads:

A car that at the initial time at rest was 2 km of a post roadman is subjected to acceleration, whose tangential component during the first 10 seconds, varies with time as shown in Fig. From that moment the brakes are applied to subject the car to an acceleration whose tangential component is constant and stops the car in the first 100 meters

a) Complete the graph shown qualitatively in Fig.
b) To obtain expressions in terms of time for the module of the velocity vector of the car, valid for each of the time intervals considered.
c) To obtain expressions in terms of time to measure the position of the car along the road, about the position roadman, valid for each of the time intervals considered.
d) Determine how far the post roadman stops the car.
e) Perform qualitative graphical function of time for the module of its velocity vector and its position, measured along the path, for the post runners.

3. The attempt at a solution

At first I did was plotted, as requested. Consider that the braking acceleration is constant, and therefore a constant line plot from the 10s.

Then I got the expressions for the velocity (from here m=meters, t=time in seconds):

$$a(t)=\displaystyle\frac{tm}{5s^3}$$ $$t\leq{10}$$

$$v(t)=\displaystyle\frac{t^2m}{10s^3}$$ $$t\leq{10}$$

$$v(10s)=10m/s$$

From here I wanted to get acceleration after ten seconds:

$$\displaystyle\frac{\Delta x}{\Delta v}=\Delta t\Rightarrow{\Delta t=10s}$$

$$a=\displaystyle\frac{\Delta v}{\Delta t}=\displaystyle\frac{-1m}{s^2}$$

$$v(t)=20\displaystyle\frac{m}{s}-\displaystyle\frac{tm}{s^2}$$

b) $$v(t)=\begin{Bmatrix} \displaystyle\frac{t^2m}{10s^3} & \mbox{ si }& t\leq{10}\\20m/s-tm/s^2 & \mbox{si}& 20s>t>10s\end{matrix}$$

c)Position $$x_i=2000m$$

Integrating I get
$$x(t)=2000m+t^3m/30s^3 \forall{t\in{[0,10]}}$$

For the initial position in the second interval I've considered the position at 10s:
$$x(10)=2000m+1000m/30=\displaystyle\frac{6100m}{3}\approx{2033.3m}$$

Then integrating:

$$x(t)=\begin{Bmatrix} x(t)=2000m+t^3m/30s^3 \forall{t\in{[0,10]}}\\\displaystyle\frac{6100m}{3}+20\displaystyle\frac{m}{s}-\displaystyle\frac{t^2m}{2s^2}\forall{t\in{[10,20]}\end{matrix}$$

This is where the problem is, the point d asks to calculate the distance that the vehicle is stopped
This distance should be according to my calculations:

$$\displaystyle\frac{6100m}{3}+100m=\displaystyle\frac{6400m}{3}$$

But this should be equal to x (t), evaluated at the 20s:

$$x(20s)=\displaystyle\frac{6100m}{3}+\displaystyle\frac{20m}{s}20s-\displaystyle\frac{(20s)^2m}{2s^2}=\displaystyle\frac{6100}{3}m+400m-200m=\displaystyle\frac{6700m}{3}\approx{2233.3m}$$

What Am I doing wrong?

Bye there!

Last edited: Aug 23, 2010
2. Aug 23, 2010