Question

From t = 0 to t = 3.51 min, a man stands still, and from t = 3.51 min to t = 7.02 min, he walks briskly in a straight line at a constant speed of 1.64 m/s. What are (a) his average velocity v_avg and (b) his average acceleration a_avg in the time interval 1.00 min to 4.51 min? What are (c) v_avg and (d) a_avg in the time interval 2.00 min to 5.51 min?

Answer 1

a) the entire interval considered is ?t = 4.51 min - 1.0 min = 3.5 min = 210 s

where as the sub interva; in which he is moving is only ?t¹ = 4.51 min - 3.51 min = 1 min = 60 s

his point at t= 1 min x = 0 at t = 4.51 min is

                     x = v?t¹

                        = 1.64m/s ( 60 s)

                        = 98.4 m

   v(avg) = (98.4 m - 0) / 210 s

               = 0.47 m/s

b) the man is at rest t = 1min and has velocity 1.64 m/s at 4.51 min thus

                  accleration ( avg) = (1.64 m/s - 0) / 210 s

                                              = 0.00781 m/s^2

c) the entire interval considered is ?t = 5.51 min - 2.0 min = 3.51 min = 210.6 s

where as the sub interva; in which he is moving is only ?t¹ = 5.51 min - 3.51 min = 2.0 min = 120 s

his point at t= 2 min x = 0 at t = 5.51 min is

                    x = v?t¹

                        = 1.64 m/s ( 120 s)

                        = 196.8 m

   v(avg) = (196.8 m - 0)/210.6 s

              = 0.9345 m/s

d)

the man is at rest t = 2min and has velocity 1.64 m/s at 5.51 min thus

                  accleration ( avg) = (1.64 m/s - 0) / 210.6 s

                                              = 0.00779 m/s^2