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

the answer is the first electromagnet (a). this is because it has more coils around the ferromagnetic material. increasing the number of coils increases the surface area in which the metal interacts with the current in the coil. this enables more atoms in the metal to become aligned hence becoming more magnetized.