Question

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.

Answer 1

Answer:

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 $a=2.41m/sec^2$

From first equation of motion $v=u+at$

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 $a=1.65m/sec^2$

So 0 =20 -1.65×t

t = 12.12 sec

So total time = 8.298 + 12.12 = 20.419 sec

So average velocity $=\frac{total\ distance}{total\ time}=\frac{1188}{20.419}=58.181m/sec$

So average velocity will be 58.181 m/sec