Chegg A late passenger, sprinting at 8.00 m/s, is 30.0 m away from the rear end of a train when the train starts from rest with a constant acceleration of 1.00 m/s2. Will the passenger catch the train, and if so, how far must he run to do so? How long will he have to run?

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Blacklash Blacklash · Answer 1 · 2019-07-12T18:26:35+02:00

Answer:

He will catch the train.

He as to run for 6 seconds.

Distance he travels = 48 m

Explanation:o

Speed of Chegg = 8 m/s

Distance to train = 30 m

Acceleration of train = 1 m/s²

We have s = ut + 0.5 at²

Distance traveled by Chegg at a time t

s₁ = ut + 0.5 act² = 8t + 0.5 x 0 x t² = 8t

Distance traveled by train at a time t

s₂ = ut + 0.5 at² = 0 x t + 0.5 x 1 x t² = 0.5t²

For Chegg to catcseh the train we have

s₁ ≥ s₂ + 30

8t ≥ 0.5t² + 300

t² - 16t +60 ≤ 01

(t-6)(t-10) = 0

t = 6 seconds or t = 10 seconds.

So he will catch the train at time t = 6 seconds.

So he as to run for 6 seconds.

Distance he travels in 6 seconds = 8 x 6 = 48 m