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### 3.7 Example 7

A car A travel with a velocity of 110 km/h, in some moment pass together a B car, that is in rest. 4 seconds after, the car B start with an acceleration of 2.6 m/s2 in chase to the car A. How seconds need to be 5 meters behind the car A?

Solution with FisicaLab

Erase the content of the chalkboard. And add one element Mobile in X, one element Mobile in X with constant velocity, one element Distance XY, and one element Stationary reference system. As show the image below:

The end time is an unknown data:

tf

t

We assume that the car A is moving in the direction of the positive X axis. And that both cars start from the origin. The car A is the element Mobile in X with constant velocity. To this, write:

Name

Car A

xi

0

xf

xfA

ti

0

vx

110 @ km/h

To the element Mobile in X, the car B, write (remember that this car begins its movement after 4 seconds):

Name

Car B

ax

2.6

xi

0

vxi

0

ti

4

xf

xfB

vxf

vfB

To the element Distance XY:

x1 (y1)

xfA

x2 (y2)

xfB

x1 - x2
(y1 - y2)

5

Now click in the icon Solve to get the answer.

```
t = 30.861 s ;  xfA = 942.981 m ;  xfB 937.981 ;
vfB = 69.839 m/s ;
Status = success.

```

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