1) A plane flies 485 km away from its base at 708 m/s, then flies back to its base at 1062 m/s. What is the average speed in m/s?
2) See attachment and solve problems 15 - 17.
1) By definition average speed = Total distance traveled/Total time taken
Total distance traveled = 2 x 485 km = 970000 m
Time taken for outwards journey = Distance/speed = 970000/708 = 1370 sec
Time taken for return journey = Distance/speed = 970000/1062 = 913 sec
Average speed = Total distance traveled/Total time taken = 970000/(1370+913) = 424.88 m/s
As the acceleration of the car changes from time to time, we have to calculate its velocity and position at different times step by step as follows :
i). At t = 0, s=0, v=0 (given)
ii) At t = 1sec. : As during the first second, the velocity and acceleration both are zero, the object continues to be at rest. Hence, at the end of the first second s=0, v=0.
iii) At t = 3 sec : initial velocity u=0, acceleration a = 3 m/s2 (from the graph), t = 2sec.
Using v = u + at, we get final velocity v = 0 ...