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A train accelerates from a station with a = 1.418m/s? Upon reaching a speed of 16.89m/s the train travels at a constant velocity for a period. The train slows down as it approaches the next station at a rate of 2m/s 2 and stops at that station Hint: Sketch a graph of uversus tfor the train's journey. (a) If the two stations are 1,200m apart, how long does the train journey take?

Respuesta :

Answer:

81.22 s

Explanation:

Period during which train accelerates  t

= [tex]\frac{16.89}{1.418}[/tex]

=11.91 s

distance travelled during acceleration

= 1/2 a t² =.5 x 1.418 x 11.91² =100.57 m

Train decelerates during the period

= [tex]\frac{16.89}{2}[/tex]

= 8.445

Distance travelled during the period of deceleration

= v² /2 a = 16.89² / 2 x 2 =71.32 m.

Distance during which train travels uniformly

= 1200 - ( 100.57 + 71.32 ) = 1028.11 m

time taken in this part of journey

= [tex]\frac{1028.11}{16.89}[/tex]

= 60.87 s

Total journey period = 60.87 + 8.445 + 11.91 =  81.22 s

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