Physics question

Question

A train on a very long stretch of tracks, starting from a dead stop, steadily speeds up to 28 m/s (about 63 mph) in 14 minutes, and then slows to 18 m/s (about 40 mph) with a constant deceleration of magnitude 0.011 m/s^2. What was the average speed of the train for this whole thing (speeding up and slowing down)? Hint: in this problem, if you take the direction the train is going as positive, the average speed is the same thing as average velocity.

Answer 1

During the acceleration step, the equation relating velocity to time is v = ((28 m/s)/((14)(60) sec)) (t)  --> v = t/30

During the deceleration step, the equation relating velocity to time is  v-28 = -.011(t-840) --> v = -.011t + 37.24

To find the average velocity, determine the total distance, then divide by the total time.  The total distance can be found using calculus:  integrate each function and evaluate the integral of the first function from 0 to 840 sec, and evaluate the integral of the second function from 840 to (840 + 10/.011) sec.

The integral of the first function is t²/60 + const.  The integral of the second function is (-.011/2)t² + 37.24t + const.

The distance covered during the acceleration step is (840²)/60 = 11760 meters

The distance covered during the deceleration step is ((840+10/.011)²/2) - 37.24(840+10/.011) + (840²/2) - 37.24(840) = 20909.090909...  meters.  The average velocity is (11760 + 20909.0909)/(840+10/.011) = 18.68 m/s

 

Another way to get the distance is just determine the area under each function curve:  The area under the first curve is represented by the area of a right triangle with a base of 840 sec and a height of 28 m/sec = 1/2(840)(28)=11760 m

The area under the second curve is represented by the area of a rectangle with length (10/.011) and a height of 18, and a right triangle with a base of (10/.011) and a height of 10.  A = 16363.6363... + 4545.4545...   = 20909.0909...