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?

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

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explanation:

there is no change in the electric potential energy as it is in rest on both conditions.