a bus travels 280 km south along a straight path with an avg velocity of 88 km/h to the south. then it travels 210 km south with an avg velocity of 75 km/h to the south.
a) how long does the 1st trip last?
b) how long does the 2nd trip last?
c) what is the avg velocity for the total trip?
t1 = 280/88
t2 = 210/75
all south so
v average south = (280+210)/(t1+t2)
To find the answers to the given questions, we can use the formula:
Time = Distance / Velocity
Let's solve each part of the question step by step:
a) How long does the 1st trip last?
In this case, we are given:
Distance (d1) = 280 km
Velocity (v1) = 88 km/h
Using the formula mentioned earlier:
Time (t1) = d1 / v1
Plugging in the values:
t1 = 280 km / 88 km/h
t1 ≈ 3.18 hours
Therefore, the first trip lasts approximately 3.18 hours.
b) How long does the 2nd trip last?
In this case, we are given:
Distance (d2) = 210 km
Velocity (v2) = 75 km/h
Using the same formula:
Time (t2) = d2 / v2
Plugging in the values:
t2 = 210 km / 75 km/h
t2 ≈ 2.8 hours
Therefore, the second trip lasts approximately 2.8 hours.
c) What is the average velocity for the total trip?
To find the average velocity for the total trip, we need to calculate the total displacement and divide it by the total time taken.
Total Displacement (D) = Distance covered south - Distance covered north
D = (d1 + d2) - 0 [Since there is no northward displacement mentioned]
Plugging in the values:
D = (280 km + 210 km) - 0
D = 490 km
Total Time (T) = t1 + t2
T = 3.18 hours + 2.8 hours
T ≈ 5.98 hours
Average Velocity (Vavg) = Total Displacement / Total Time
Vavg = D / T
Plugging in the values:
Vavg = 490 km / 5.98 hours
Vavg ≈ 81.9 km/h
Therefore, the average velocity for the total trip is approximately 81.9 km/h.