Kinematics Equation

Question

A ball rolling down a hill was displace 5 m while uniformly accelerating from rest. If the final velocity was 20 m/s. What was the rate of acceleration?

Your answer

Answer 1

Given: x_f - x₀ = 5 m

v₀ = 0 m/s

v_f = 20 m/s

a = constant

v_f² = v₀² + 2a(x_f - x₀)

a = (v_f² - v₀²) / 2(x_f - x₀) = ((20 m/s)² - (0 m/s)²) / 2(5 m) = 40 m/s²

Answer 2

Use the equation

vf² = vi² + 2 * a * (xf - xi)

final velocity vf = 20m/s

initial velocity vi = 0m/s

final position xf = 5 m

initial position xi = 0 m

Use these to find a.

20² = 0 + 2 * a * (5 - 0)

400 = 10a

a = 40m/s²

Answer 3

Since you know the displacement (5m), initial velocity (it's at rest, so initial velocity is 0), and final velocity (20 m/s), you can use the following equation to find acceleration:

V_f² = V_i² + 2ad

Substituting the values into the equation we get

20² = 0² * 2a*5

400 = 10a

40 = a

So the acceleration is 40 ^m/_s²