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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.

Previous: Example 6 (k), Up: Examples kinematics of particles [Contents][Index]