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Tomtit [17]
3 years ago
5

A train moving at a constant speed of 50.0 km/h moves east for 45.0 min, then in a direction 45.0° east of due north for 10.0 mi

n, and then west for 55.0 min. What is the average velocity of the train during this run? (Let the +x-axis point towards the east and the +y-axis point towards the north.)l

Physics
1 answer:
Yuki888 [10]3 years ago
8 0

Answer:

3.47km/h with a direction of 67.8 degrees north of west.

Explanation:

First we need to calculate the displacement on the X axis, so:

d_{x1}=50.0km/h*45min(\frac{1h}{60min})=37.5km\\d_{x2}=50.0km/h*10min(\frac{1h}{60min})*cos(45^o)=5.90km\\d_{x1}=-50.0km/h*55.0min(\frac{1h}{60min})=45.8km\\D_x=37.5+5.90-45.8=-2.4km

then on the Y axis:

D_y=50.0km/h*10min(\frac{1h}{60min})*sin(45^o)=5.90km

The magnitud of the displacement is given by:

D=\sqrt{D_x^2+D_y^2} \\D=6.37km

and the angle:

\alpha =arctg(\frac{5.90}{2.41})=67.8^o

that is 67.8 degrees north of west.

v=\frac{D}{t}\\v=\frac{6.37km}{110min*\frac{1h}{60min}}\\v=3.47km/h

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Vinvika [58]
<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}    (1)

V=V_{o}-g.t    (2)

Where:

y  is the rock's final position

y_{o}=0  is the rock's initial position

V_{o}=30\frac{m}{s} is the rock's initial velocity

V is the final velocity

t is the time the parabolic movement lasts

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As we know y_{o}=0 , equation (2) is rewritten as:

y=V_{o}.t+\frac{1}{2}g.t^{2}    (3)

On the other hand, the maximum height  is accomplished when V=0:

V=V_{o}-g.t=0    (4)

V_{o}-g.t=0    

V_{o}=g.t    (5)

Finding t:

t=\frac{V_{o}}{g}    (6)

Substituting (6) in (3):

y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}    (7)

y_{max}=\frac{{V_{o}}^{2}}{2g}    (8)  Now we can calculate the maximum height of the rock

y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}   (9)

Finally:

y_{max}=277.777m  

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