A train starts from rest and accelerates uniformly until it has traveled 5.6 km and acquired a forward velocity of The train then moves at a constant velocity of for 420 s. The train then slows down uniformly at until it is brought to a halt. The acceleration during the first5.6 km of travel is closest to: a) 0.19 m/s2
b) 0.14 m/s2
c) 0.16 m/s2
d) 0.20 m/s2
e) 0.17 m/s2

c) 0.16 m/s2

Explanation:

The computation of the acceleration during the first km of travel is shown below

Given that

Final velocity = v = 42 m/s

Initial velocity = u = 0 m/s

Distance = 5.6km

Based on the above information, we need to apply the following formula

As we know that

= 0.1575 m/s ^2

hence, the correct option is c. 0.16 m.s^2

50 inches should be the resultant vector

gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.