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tia_tia [17]
2 years ago
10

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? (It is not necessary to convert your answer from kilometers to meters for this problem. You may leave it in kilometers.)
Physics
1 answer:
sergejj [24]2 years ago
5 0

Answer:

0.176m from the flagpole, westward.

Explanation:

Let the Eastward be the positive direction. So initially runner A is at position -6km, running with velocity of 9km/h while runner B is at position 5km running at a velocity of -8km/h. We can conduct the following equation for their distances over the same time t

s_A = -6 + 9t

s_B = 5 - 8t

When A an B meets, they are at the same position and at the same time. So

s_A = s_B

-6 +9t = 5 - 8t

17t = 5 + 6 = 11

t = 11/17 = 0.647 s

s_A = -6 + 9*0.647 = -0.176 m

So where they meet is 0.176m from the flagpole, westward.

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Answer:

Blue Lighting

Explanation:

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2 years ago
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A train started from rest and moved with constant acceleration. At one time it was traveling 27 m/s, and 150 m farther on it was
AlekseyPX

Explanation:

(a) Given:

Δx = 150 m

v₀ = 27 m/s

v = 54 m/s

Find: a

v² = v₀² + 2aΔx

(54 m/s)² = (27 m/s)² + 2a (150 m)

a = 7.29 m/s²

(b) Given:

Δx = 150 m

v₀ = 0 m/s

a = 7.29 m/s²

Find: t

Δx = v₀ t + ½ at²

150 m = (0 m/s) t + ½ (7.29 m/s²) t²

t = 6.42 s

(c) Given:

v₀ = 0 m/s

v = 27 m/s

a = 7.29 m/s²

Find: t

v = at + v₀

27 m/s = (7.29 m/s²) t + 0 m/s

t = 3.70 s

(d) Given:

v₀ = 0 m/s

v = 27 m/s

a = 7.29 m/s²

Find: Δx

v² = v₀² + 2aΔx

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7 0
2 years ago
Let’s look at a radio-controlled model car. Suppose that at time t1=2.0st1=2.0s the car has components of velocity vx=1.0m/svx=1
Nonamiya [84]

Answer:

a_x=6\ \text{m/s}^2 and a_y=0\ \text{m/s}^2

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Explanation:

t_1=2\ \text{s}

v_x=1\ \text{m/s}

v_y=3\ \text{m/s}

t_2=2.5\ \text{s}

v_x=4\ \text{m/s}

v_y=3\ \text{m/s}

Average acceleration in the different axes

a_x=\dfrac{\Delta v_x}{\Delta t}\\\Rightarrow a_x=\dfrac{4-1}{2.5-2}\\\Rightarrow a_x=6\ \text{m/s}^2

a_y=\dfrac{\Delta v_y}{\Delta t}\\\Rightarrow a_y=\dfrac{3-3}{2.5-2}\\\Rightarrow a_y=0\ \text{m/s}^2

The components of the acceleration is a_x=6\ \text{m/s}^2 and a_y=0\ \text{m/s}^2

The magnitude of acceleration

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Direction

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The magnitude of accleration is 6\ \text{m/s}^2 and the direction is 0^{\circ}.

7 0
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