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weeeeeb [17]
3 years ago
9

A train travels from Albuquerque, New Mexico, to Mexico City in 4 hours with an average velocity of 80 km/h to the south. What i

s the train’s displacement?
Physics
2 answers:
Ivenika [448]3 years ago
6 0
The displacement of the train is equal to the velocity of the train multiplied by the time of travel. In this case,the average velocity is equal to 80 km/h and the time of travel is four hours. The answer is 80 km/h *4 h or equal to 320 km in total. Answer is 320 kilometers.
nikitadnepr [17]3 years ago
6 0

Answer: The displacement is the difference in position (between the final position and the initial), so train's displacement will be the initial position subtracted of the position when the train has traveled for 4 hours.

If we define the original position as 0, then the total displacement is, if the train has a velocity of 80km/h, then in 4 hours the train travels 80*4 km= 320 km to the south. Because the train travels in one direction and never turns, the total displacement is 320km

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Answer:

123.30 m

Explanation:

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Speed, u = 22 m/s

acceleration, a = 1.40 m/s²

time, t = 7.30 s

From equation of motion,

                       v = u + at

where,

v is the final velocity

u is the initial velocity

a is the acceleration

t is time  

                       V = at + U

using equation  v - u = at to get line equation for the graph of the motion of the train on the incline plane

                       V_{x} = mt + V_{o}      where m is the slope

Comparing equation (1) and (2)

V = V_{x}

a = m    

U = V_{o}

Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.

          a = -  1.40 m/s²

Sunstituting a = -  1.40 m/s² and  u = 22 m/s

                        V_{x} = -1.40t + 22

                            V_{x} = -1.40(7.30) + 22

                             V_{x} = -10.22 + 22

                             V_{x} = 11. 78 m/s

The speed of the train at 7.30 s is 11.78 m/s.

The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.

           Area of triangle +  Area of rectangle

          [\frac{1}{2} * (22 - 11.78) * (7.30)]  + [(11.78 - 0) * (7.30)]

                           = 37.303 + 85.994

                           = 123. 297 m

                           ≈ 123. 30 m

                 

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