Respuesta :
Answer:
Part a)98 km/h
Part b) 93.5 km/h
Part c) 95.7 km/h
Explanation:
A) For the trip from Antonio to Houston we have
Let the total time in which the trip is completed be 't'
Thus the distance covered with a speed of 77 km/h equals [tex]77\times \frac{t}{2}[/tex]
Similarly the distance covered with a speed of 119 km/h equals [tex]119\times \frac{t}{2}[/tex]
Now by definition of average speed we have
Average Speed = [tex]\frac{Distance}{time}[/tex]
Applying values we get
Average Speed for trip 1 = [tex]\frac{77\times \frac{t}{2}+119\times \frac{t}{2}}{t}[/tex]
Thus average speed equals [tex]\frac{77+119}{2}=98km/h[/tex]
B)For the trip from Houston Antonio to we have
Let the total distance of the trip be 'd'
Time required to complete half the distance at a speed of 77 km/h equals [tex]t_{1}=\frac{0.5d}{77}[/tex]
Time required to complete remaining half the distance at a speed of 119 km/h equals [tex]t_{1}=\frac{0.5d}{119}[/tex]
Now by definition of average speed we have
Average Speed = [tex]\frac{Distance}{time}[/tex]
Applying values we get
Average Speed for trip 1 = [tex]\frac{d}{\frac{0.5d}{77}+\frac{0.5d}{119}}[/tex]
Thus average speed equals [tex]\frac{1}{\frac{0.5}{77}+\frac{0.5}{119}}=93.5km/h[/tex]
C) For the entire trip we have
Average Speed = [tex]\frac{Distance}{time}[/tex]
Applying values we get
Average Speed for both the trips = [tex]\frac{2d}{\frac{2d}{77+119}+\frac{0.5d}{77}+\frac{0.5d}{119}}[/tex]
Thus average speed for both the trips equals 95.7 km/h
