Question

What is the acceleration of a ball rolling down a ramp that starts from rest and travels 0.9 m in 3 s?

Answer 1

The acceleration of the ball rolling down the ramp is 0.2 m/s^2, calculated using the equation of motion for uniformly accelerated movement without initial velocity.

Step-by-step explanation:

To calculate the acceleration of a ball that starts from rest and travels a certain distance down a ramp over a known time period, we can use the equations of motion for uniformly accelerated motion. It is given that the ball travels 0.9 meters in 3 seconds from rest.

The equation that relates distance (s), initial velocity (u), time (t), and acceleration (a) is:

s = ut + \frac{1}{2}at^2

Since the initial velocity u is 0 m/s (because the ball starts from rest), the equation simplifies to:

s = \frac{1}{2}at^2

Rearranging this equation to solve for acceleration yields:

a = \frac{2s}{t^2}

Plugging in the given values:

a = \frac{2 * 0.9 m}{(3 s)^2} = \frac{1.8 m}{9 s^2} = 0.2 m/s^2

Therefore, the acceleration of the ball is 0.2 m/s2.

Answer 2

Given:
u = 0, initial velocity
s 0.9 m, distance traveled.
t = 3 s, the time taken.

Let a = the acceleration. Then
s = ut + (1/2)*a*t²
(0.9 m) = 0.5*(a m/s²)*(3 s)²
0.9 = 4.5a
a = 0.2 m/s²

Answer: 0.2 m/s²