Arthur and Betty start walking toward each other when they are 157 m apart. Arth

Question

Arthur and Betty start walking toward each other when they are 157 m apart. Arthur has a speed of 4.3 m/s and Betty has a speed of 3.2 m/s. Their dog, Spot, starts by Arthur's side at the same time and runs back and forth between them at 8 m/s. By the time Arthur and Betty meet, what distance, in meter, has Spot run? The position of an object is given as a function of time as x(t) = (-1.92 m/s)t + (6.8 m/s^2)t^2. What is the average speed, in m/s, of the object between t = 0.79 s and t = 4.45 s?

Explanation / Answer

time taken to meet = d/relative speed = 157/(4.3+3.2) = 20.9 s

distance travelled = 8*20.9 = 167.2 m

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at t = 0.79

x1 = (-1.92*0.79)+(6.8*0.79^2) = 2.73m

at t= 4.45 s

x2 = (-1.92*4.45)+(6.8*4.45^2) = 126.113 m

speed = (x2-x1)/(t2-t1) = (126.113-2.73)/(4.45-0.79) = 33.7 m/s

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