Starting from rest, an object rolls down a long ramp with a constant acceleration of 1.72 m/s2. After rolling for 4.75 s, what is the magnitude of the ball’s final velocity?

A. 9.81 m/s
B. 8.17 m/s
C. 6.30 m/s
D. 4.09 m/s

Well, this is quite the rollin' situation we have here! Let's solve it with some humor, shall we?

So, the object is rolling down a ramp with a constant acceleration of 1.72 m/s². After 4.75 seconds, we want to find the magnitude of its final velocity.

To solve this, we need to use the formula for final velocity: vf = vi + at, where vf is the final velocity, vi is the initial velocity (in this case, 0 m/s since it starts from rest), a is the acceleration (1.72 m/s²), and t is the time (4.75 s).

Plugging in the values, we get: vf = 0 + (1.72 m/s²)(4.75 s) = 8.17 m/s.

So, the answer is B. 8.17 m/s. The object is rolling towards the finish line with quite the speed!

To find the magnitude of the ball's final velocity, we can use the equation:

vf = vi + at

where vf is the final velocity, vi is the initial velocity (which is 0 because the object starts from rest), a is the acceleration, and t is the time.

Plugging in the values we know:

vf = 0 + (1.72 m/s^2)(4.75 s)

Calculating the value:

vf = 1.72 m/s^2 * 4.75 s

vf = 8.17 m/s

Therefore, the magnitude of the ball's final velocity is 8.17 m/s.

The correct answer is B. 8.17 m/s.

To find the magnitude of the ball's final velocity, we can use the constant acceleration kinematic equation:

v = u + at

Where:
v = final velocity
u = initial velocity (in this case, the object starts from rest, so u = 0)
a = acceleration (given as 1.72 m/s^2)
t = time (given as 4.75 s)

Substituting the values into the equation, we get:

v = 0 + (1.72 m/s^2)(4.75 s)

v = 8.17 m/s

Therefore, the magnitude of the ball's final velocity is 8.17 m/s.

The correct answer is B. 8.17 m/s.

v = a t

hey, try some of these yourself !