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would anyone be able to help me? I'm struggling with the last 2 parts

Would Anyone Be Able To Help Me Im Struggling With The Last 2 Parts class=

Sagot :

onyebuchinnaji

The time taken for the train to stop is 8.9 seconds.

The distance traveled by the train during acceleration is 9,984 m.

The distance traveled by the train during deceleration is 17.8 m.

Time taken for the train to stop

Use the following kinematic equation to determine the time taken for the train to stop.

v = u + at

where;

  • v is the final velocity of the train = 0
  • u is the initial velocity = 4 m/s
  • a is the deceleration = -0.45 m/s²
  • t is time of motion

0 = 4 - 0.45t

0.45t = 4

t = 4/0.45

t = 8.89 seconds = 8.9 seconds

Distance traveled by the train during acceleration

s = ut + ¹/₂at²

where;

  • t is time of motion = 8 min = 480 seconds

s = (4)(480) +  ¹/₂(0.07)(480)²

s = 9,984 m

Distance traveled by the train during deceleration

s = ut + ¹/₂at²

s = (4 x 8.9) + ¹/₂(-0.45)(8.9)²

s = (4 x 8.9) - ¹/₂(0.45)(8.9)²

s = 17.8 m

Learn more about deceleration here: https://brainly.com/question/75351

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