a) If it accelerate at the rate of 10ft/s^2, starting from rest at the time t=0,

Its speed is `10t` ft/s

The distance is the anti derivative of the velocity

`d= 10t^2/2=5t^2 ft.`

The maximum speed is 120 mi/h ie 176 ft/s

The train reaches its maximum speed at...

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a) If it accelerate at the rate of 10ft/s^2, starting from rest at the time t=0,

Its speed is `10t` ft/s

The distance is the anti derivative of the velocity

`d= 10t^2/2=5t^2 ft.`

The maximum speed is 120 mi/h ie 176 ft/s

The train reaches its maximum speed at `t=176/10=17.6s`

The distance during the acceleration period is `d=5*17.6^2=1548.8ft`

Then the train travels for 15mn ( a quarter of an hour) at the speed of 120mi/h.

The distance is `120/4=30mi.`

1548.8 ft is 0.293mi

**Therefore the maximum total distance is** 30.293mi.