Question

the velocity of a moving object, v, in meters per second, is given by the formula below. The variable is the kinetic energy of the object, in joules, and m is the mass of the object, in kilograms.

A marble with a mass of 0.04 kilograms rolls down a ramp and reaches a maximum kinetic energy of 0.875 Joules. Use the given formula to write a function, v(x)c that models the situation. Then use the function to predict how the velocity of the marble changes as it rolls down the ramp

Answer 1

Assuming the velocity is as written,
a = 3(2) = 6

If you meant 3(2t+1)⁰.⁴, then
a = 2.4/(2t+1)⁰.⁶

as usual,
s = 25 + 1/2 at²
= 25 + 3t² 
or
= 25 + 1.2(2t+1)¹.⁴

Answer 2

Answer: Option B: As the kinetic energy increases, velocity increases until it reached the maximum kinetic energy.

Explanation:

From law of conservation of energy, total energy of a closed system remains constant. One form of energy can convert to another.

Here, when a marble is kept on the ramp, it would have energy = potential energy. On reaching the bottom of the ramp, all the potential energy would be converted to kinetic energy (maximum). In between, it would have both potential as well as kinetic energy.

$K.E. (max) = \frac{1}{2}mv_{max}^2$

Potential energy at any given point on ramp, = mg x where x is the height of the marble which varies as marble rolls down the ramp.

$\frac{1}{2}mv^2 = \frac{1}{2}mv_{max}^2- mg x$

$v = \sqrt{v_{max}^2- 2g x}$

So, as the kinetic energy increases, the velocity increases until it reaches the maximum kinetic energy. correct option is B.