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AfilCa [17]
4 years ago
13

Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 9.0 km/h due east. Runner B is initia

lly 5.0 km east of the flagpole and is running with a constant velocity of 8.0 km/h due west. What will be the distance of the two runners from the flagpole when their paths cross
Physics
1 answer:
Eduardwww [97]4 years ago
7 0

Answer:

From the data we know that runner A and runner B are 11 km apart from the start because (6+5) km

So the runner from the east direction has distance as unknown km, rate= 9 k/h ; time= d/r=x/9 hr

So runner towards the west will be

distance = 11-x, rate= 8 k/h, time = d/r = (11-x)/8

So equating east and west time we have

x/9= (11-x)/8

8x=99-9x

17x=99

x=5.92 km

That is the distance covered by runner towards the east and he will meet the runner toward the west at

6-5.92=0.08 km west of the flagpole.

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Answer:

\displaystyle \vec{d}=

Explanation:

<u>Displacement Vector</u>

Suppose an object is located at a position  

\displaystyle P_1(x_1,y_1)

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\displaystyle P_2(x_2,y_2)

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\displaystyle \vec{d}=

If the position is given as magnitude-angle data ( z , α), we can compute its rectangular components as

\displaystyle x=z\ cos\alpha

\displaystyle y=z\ sin\alpha

The question describes the situation where the initial point is the base of the mountain, where both components are zero

\displaystyle P_1(0,0)

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\displaystyle x_2=520\ cos32^o= 440.99\ m

\displaystyle y_2=520\ sin32^o=275.6\ m

The displacement is

\displaystyle \vec{d}=

5 0
3 years ago
julia throws a ball vertically upward from the ground with a speed of 5.89m/s. Andrew catches it when it is on its way down at a
aliya0001 [1]
Vo = 5.89 m/s Y = 1.27 m g = 9.81 m/s^2 
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Tr = Vo / g Tr = (5.89 m/s) / (9.81 m/s^2) Tr = 0.60 s 
Max height achieved is:
H = Vo^2 / [2g] H = (5.89 )^2 / [ 2 * (9.81) ] H = (34.69) / [19.62] H = 1.77 m 
It falls that distance, minus Andrew's catch distance:
h = H - Y h = (1.77 m) - (1.27 m) h = 0.5 m 
Time to descend is therefore:
Tf = √ { [2h] / g ] Tf = √ { [ 2 * (0.5 m) ] / (9.81 m/s^2) } Tf = √ { [ 1.0 m ] / (9.81 m/s^2) } Tf = √ { 0.102 s^2 } Tf = 0.32 s 
Total time is rise plus fall therefore:
Tt = Tr + Tf Tt = (0.60 s) + (0.32 s) Tt = 0.92 s           (ANSWER)
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Answer:

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