To find the displacement of the bank robber during this time interval, we need to calculate the distance he traveled in each direction and then find the vector difference between his initial and final positions.
From 3:00 P.M. to 3:37 P.M., the bank robber traveled for 37 minutes = 0.617 hours at a speed of 135.0 mi/h. Therefore, he traveled a distance of:
d1 = v1 x t1 = 135.0 mi/h x 0.617 h = 83.3 miles north
From 3:00 P.M. to 3:37 P.M., the bank robber traveled for 37 minutes = 0.617 hours at an average speed of 103.0 mi/h. Therefore, he traveled a distance of:
d2 = v2 x t2 = 103.0 mi/h x 0.617 h = 63.5 miles north
To find the displacement, we need to subtract the initial position from the final position. The initial position is at milepost 129 and the final position is at milepost 182. Therefore, the displacement is:
displacement = final position - initial position
displacement = 182 - 129 = 53 miles north
Since the displacement is in the positive direction, we enter a positive value. Therefore, the bank robber's displacement during this time interval is 53 miles north.