Atrain starts from rest and accelerates uniformly, until ithas traveled 5.6 km and acquired a velocity of 42 m/s. thetrain then moves at a constant velocity of 42 m/s for 420 s. the train then slows down uniformly at 0.065 m/s2, untilit is brought to a hault. the acceleration during the first5.6 km of travel is closest to: a) 0.19 m/s2b) 0.14 m/s2c) 0.16 m/s2d) 0.20 m/s2e) 0.17 m/s2

c

Explanation:

Hello!

To solve this problem we need to know the velocity and position of the train as a fucntion of time and acceleration. Since the train started from rest, these are the formulas:

v = at

x = (1/2)a*t^2

From the first equation we can know the time as a division between the velocity and the acceleration:

t = (v/a)

Replacing this value in the second equation we get:

x = (1/2)*a*(v/a)^2 = v^2/(2*a)

Solving for a:

a = (1/2)*(v^2/x)

Now, we know that when x=5.6km =5600 m, the velocity of the train is v =42 m/s

Therefore:

a = (1/2)*(42^2/5600) m/s^2 = 0.1575 m/s^2

So, the answer is c, the acceleration of  the train during the first 5.6 km is colse to 0.16m/S^2

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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c

explanation: