An object at rest near the surface of a distant planet starts to fall freely. If the acceleration there is twice that of the Earth, its speed one second later would be
A)10 m/s.
B)20 m/s.
C)30 m/s.
D)40 m/s.

Question

Quizplus · Accepted Answer

According to the equation of motion, v = u + at, where v is the final velocity, u is the initial velocity (which is zero in this case), a is the acceleration and t is the time taken.

On this distant planet, the acceleration = 2 * acceleration due to gravity on Earth = 2 * 9.8 m/s^2 = 19.6 m/s^2.

Plugging the values into the equation, we get:

v = 0 + (19.6 m/s^2 * 1 s) = 19.6 m/s

Therefore, the object's speed one second later would be 19.6 m/s, which is closest to option B) 20 m/s.

An Object at Rest near the Surface of a Distant