Give your answer in si units and to three significant figures. a train departs station a to travel to station b which is 1188 meters away. the train begins at rest and accelerates at a rate of 2.41 m/s^2 until it reaches a speed of 20.0 m/s. as the train approaches station b, it begins to decelerate at a rate of 1.65 m/s^2 so that it comes to a complete stop just as it arrives at station b. determine the average velocity of the train over the entire trip.

Average velocity will be 58.181 m/sec      

Explanation:

Distance between station A and station B = 1188 meters

As the train starts from station A its initial velocity u = 0 m/sec

Final velocity is when it reaches at station B is 20 m/sec

Acceleration

From first equation of motion

20 = 0+2.41×t

t = 8.298 sec

Now from station train began to deaccelerate and finaly stop so final velocity v = 0 m /sec

Initial velocity u = 20 m/sec

We know that v = u+at

Deacceleration

So 0 =20 -1.65×t

t = 12.12 sec

So total time = 8.298 + 12.12 = 20.419 sec

So average velocity

So average velocity will be 58.181 m/sec

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