6.a train starts from rest and accelerates uniformly until it has traveled 5.6 km and acquired a velocity of 42m/s what is the approximate acceleration of the train during this time?
0.17 m/s2
0.20 m/s2
0.16 m/s2
0.14 m/s2
0.19 m/s2

0.16 m/s²

From one of the equation of linear motion;

vf2 - vi2 = 2ax

Where;

vf = final speed after traveling a distance x at constant acceleration; vf = 42 m/s

vi = initial speed = 0 (starts from rest)

a = constant acceleration while traveling a distance x

x = distance traveled at constant acceleration; x = 5.6 km = 5600 m

Therefore;

a = vf2/(2x)

  = 422/(2*5600) m/s²

  = 0.1575 m/s²

   ≅ 0.16 m/s²

battery with hi vtge high val resist in series

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft