A student releases a marble from the top of a ramp. The marble increases speed while on the ramp then continues across the floor. The marble
travels a total of 140cm in 4.10s.
What was the marble's final speed?

Help please

Vf = 0.68 m/s

Explanation:

First we find acceleration of the marble by using 2nd equation of motion:

s = Vi t + (1/2)at²

where,

s = distance covered = 140 cm = 1.4 m

t = time taken = 4.1 s

Vi = initial Velocity = 0 m/s (since, the marble was initially at rest)

a = acceleration = ?

Therefore,

1.4 m = (0 m/s)(4.1 s) + (1/2)(a)(4.1 s)²

a = (1.4 m)(2)/(4.1 s)²

a = 0.17 m/s²

Now, we use 1st equation of motion for final velocity (Vf)

Vf = Vi + at

Therefore,

Vf = 0 m/s + (0.17 m/s²)(4.1 s)

Vf = 0.68 m/s

the total distance traveled by the car is

30km + 49km = 79 km .

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.