Question

1. An object is moving at 4.66 m/s when it accelerates at 5.66 m/s2 for a period of 2.35 s. The distance it moves in this time is ____ m.

2. A Star fleet science officer drops a large rock from a height of 1.55 m onto
the surface of an alien planet. The rock strikes the ground in 0.230 s.
Determine the acceleration of gravity.

Answer 1

1. Distance, S = 26.58 meters

2. Acceleration of gravity, g = 58.60 m/s²

Step-by-step explanation:

1. Given the following data;

Initial velocity = 4.66 m/s

Acceleration = 5.66 m/s²

Time = 2.35 seconds

To find the distance travelled by the object, we would use the second equation of motion;

S = ut + ½at²

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

S = 4.66*2.35 + ½*5.66*2.35²

S = 10.951 + (2.83 * 5.5225)

S = 10.951 + 15.629

S = 26.58 meters

2. Given the following data;

Displacement (height) = 1.55 m

Time, t = 0.23 seconds

To find the acceleration of gravity, we would use the following formula;

Height = ½gt²

Substituting into the formula, we have;

1.55 = ½ * g * 0.23²

1.55 = ½ * g * 0.0529

1.55 = 0.02645g

Acceleration of gravity, g = 1.55/0.02645

Acceleration of gravity, g = 58.60 m/s²