Assume a maximum stopping acceleration of g/2 where g is acceleration due to gravity.
Answer:
2.99 m/s
Explanation:
Stopping distance, s = 3 ft = 0.914 m
final velocity, v = 0
a = g/2 = 4.9 m/s²
Use third equation of motion:
substitute the values to find the speed of train:
Answer:
boy if you dont get t f outa here
As we know that total work done by a force is given by
so it is product of force and displacement along same direction
as we can write it as
so it must be the product of force and displacement in same directions so correct answer must be
<u>B. in the same direction as the displacement vector and the motion.</u>