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You are rushing to the train station to catch your morning commute. The train leaves the train station from rest with an acceleration of 0.6 m/s^2. You arrive at the station exactly 4.0 seconds after the train leaves and you immediately start running after the train with a constant velocity of 8.5 m/s.
a. How long after the train leaves the station do you catch up with the train?
b. How far from the train station do you catch up with the train?
c. With what minimum speed would you have to run in order to catch up with the train?

Respuesta :

Answer:

a) t= 4.81 s,and t= 23.51.

b)d=6.88 m

c)v=4.8 m/s

Explanation:

Acceleration of train ,a= 0.6 m/s²

u = 0 m/s

Your speed ,V= 8.5 m/s

Lets take after t time he you catch the train

Distance travel by train in t time

[tex]d=ut+\dfrac{1}{2}at^2[/tex]

[tex]d=\dfrac{1}{2}\times 0.6\times t^2[/tex]  ----------1

d= V ( t- 4)

d= 8.5 ( t- 4)  --------2

By equating equation 1 and 2

0.3 t² =  8.5 ( t- 4)

0.3 t² -8.5 t + 34 = 0

t= 4.81 s and t= 23.51

It means that you will catch train after t= 4.81 s,and t= 23.51.

t=23.51 sec means that you will catch the train after 23.51 sec also because acceleration of train is low.

Distance travel

d= 8.5 ( t- 4)

t= 4.81 s

d= 8.5 ( 4.81- 4)  m

d=6.88 m

Lets speed = v

0.3 t² =  v ( t- 4)

0.3 t² - v t + 4 v = 0

To have one solution

[tex]D=\sqrt{b^2-4ac}[/tex]

D= 0 Should be zero.

v²- 4 x 0.3 x 4 v

v = 4 x 0.3 x 4

v=4.8 m/s

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