A car travels along a straight line at a constant speed of 58.5 mi/h for a dista

Question

A car travels along a straight line at a constant speed of 58.5 mi/h for a distance d and then another distance d in the same direction at another constant speed. The average velocity for the entire trip is 33.0 mi/h. (a) What is the constant speed with which the car moved during the second distance d? mi/h (b) Suppose the second distance d were traveled in the opposite direction; you forgot something and had to return home at the same constant speed as found in part (a). What is the average velocity for this trip? mi/h (c) What is the average speed for this new trip? mi/h pleease show work

Explanation / Answer

a) E1: v1=43.0 mi/h E2: d = v1*t1 [1st distance, constant speed v1] E3: d = v2*t2 [2nd distance, another constant speed v2] E4: v3=33.5 mi/h E5: 2d = v3*(t1+t2) [double distance, average velocity v3] Combine E2 and E3 and calculate t2 v2*t2 = v1*t1 E6: t2 = v1*t1/v2 In E5, substitute t2 with t2 from E6 and d with d from E2 2*v1*t1 = v3*(t1 + v1*t1/v2) 2*v1*t1 = v3*t1 + v3*v1*t1/v2 2*v1*t1 - v3*t1 = v3*v1*t1/v2 (2*v1-v3)*t1 = v3*v1*t1/v2 2*v1-v3 = v3*v1/v2 or E7: v2 = v3*v1/(2*v1-v3) In E7 substitute v1 with v1 from E1 and v3 with v3 from E4 v2 = 33.5*43 / (2*43 - 33.5) v2 = 27.438095 mi/h =============== b) There are 2 options: b1) if you take distance d as a "scalar" then you traveled d+d = 2d and the velocity is the same as in a) b2) if you take distance d as a "vector" then you traveled d-d = 0 and the velocity is 0 =====

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