+79 Would appreciate if someone could shed some light on how to do these questions step by step, Thank you very much..
1)A child and his father are 89m apart. They start walking towards each other. A mosquito that flies at 5.1m/s moves back and forth between the child and his father until they meet. If the mosquito started off from the same position as the child, and given that the child walks at 1.2m/s, while the father is twice as fast, what was the total distance covered by the mosquito? Keep in mind that the mosquito, child and the father start moving at the same time.
2)A subway moves between two stations. The first third of the distance is covered in uniform motion, at 20m/s. During the second third, the velocity of the train is maintained constant at 20km/h [N]. For the rest of the trip, the train was in uniform motion and the speed was 50km/h. If the train was moving towards north for the whole trip, what was the average speed of the train?
3)How fast does an arrow must be shot straight in the air, in order to cover 22.3m during its 2nd one-second time interval? Use gravitational acceleration as 9.8m/s^2.
+130071
Not great at Physics......but here's my best shot......!!!!
1)A child and his father are 89m apart. They start walking towards each other. A mosquito that flies at 5.1m/s moves back and forth between the child and his father until they meet. If the mosquito started off from the same position as the child, and given that the child walks at 1.2m/s, while the father is twice as fast, what was the total distance covered by the mosquito? Keep in mind that the mosquito, child and the father start moving at the same time.
Note that the child walks at 1.2m/s and the father walks twice as fast = 2.4 m/s.....so....the distance that they close between themselves every second must be (1.2 + 2.4) m/s = 3.6 m/s
So......it will take them 89m / 3.6 (m/s) = (445 /18 ) seconds in which to meet each other
Since the mosquito is in the air the same amount of time...it must cover a distance of (445/18)s * 5.1 m/s ≈ 126.1 m
2)A subway moves between two stations. The first third of the distance is covered in uniform motion, at 20m/s. During the second third, the velocity of the train is maintained constant at 20km/h [N]. For the rest of the trip, the train was in uniform motion and the speed was 50km/h. If the train was moving towards north for the whole trip, what was the average speed of the train?
Convert km/h to m/s
20 km/h = 20 * 1000 / 3600 = 50/9 m/s
50 km/h = 50 * 1000 / 3600 = 125/9 m/s
Note that distance / rate = time.....call the total distance that the train traveled, D
So....the time of the first leg of the trip can be denoted as (1/3)D / 20 m/s = (1/60)D = D/60
And the time of the second leg of the trip can be denoted as (1/3)D / (50/9)m/s = (3/50)D = 3D/50
And the time of the third leg of the trip can be denoted as (1/3)D / (125/9)m/s =(3/125)D = 3D/125
So........we have that
Total Distance / Total Time = Average Rate .....so......
D / [ D/60 + 3D/50 + 3D / 125] which can be factored as
D / [ D ( 1/60 + 3/50 + 3 / 125) ] =
1 / [ 1/60 + 3/50 + 3/125] =
1 / [ (25 + 90 + 36) / 1500 ] =
1 / [151 / 1500 ] =
1500 / 151 ≈ 9.9 m/s = average rate
3)How fast does an arrow must be shot straight in the air, in order to cover 22.3m during its 2nd one-second time interval? Use gravitational acceleration as 9.8m/s^2.
Using
y = V * t - (1/2) (9.8) * t^2
Where y is the height after some t seconds and V is the initial velocity
After the first second the arrow has reached a height of y1
y1 = V(1) - (1/2)(9.8) (1)^2
y1 = V - 4.9 (1)
After the next second the arrow has reached a height of y1 + 22.3m
y1 + 22.3 = V * (2) - (1/2) (9.8) * (2)^2
y1 + 22.3 = 2 V - 2 (9.8)
y1 = 2V - 2(9.8) - 22.3
y1 = 2V - 41.9 (2)
Equate (1) and (2)
V - 4.9 = 2V - 41.9 add 41.9 to both sides subtract V from both sides
37 = V = 37 m/s = the initial velocity
