Question

An object starts from rest at time t = 0.00 s and moves in the +x direction with constant acceleration. The object travels 14.0 m from time t = 1.00 s to time t = 2.00 s. What is the acceleration of the object?a) 5.20 m/s^2 b) 10.4 m/s^2 c) 8.67 m/s^2 d) 6.93 m/s^2 e) 12.1 m/s^2

Answer 1

Answer:

The acceleration of the object is 9.3 m/s²

Explanation:

For a straight movement with constant acceleration, this equation for the position applies:

x = x0 + v0 t + 1/2 a t²

where

x = position at time t

x0 = initial position

v0 = initial velocity

a = acceleration

t = time

we have two positions: one at time t = 1 s and one at time t = 2 s. We know that the difference between these positions is 14.0 m. These are the equations we can use to obtain the acceleration:

x₁ = x0 + v0 t + 1/2 a (1 s)²

x₂ = x0 + v0 t + 1/2 a (2 s)²

x₂ - x₁ = 14 m

we know that the object starts from rest, so v0 = 0

substracting both equations of position we will get:

x₂ - x₁ = 14

x0 + v0 t + 1/2 a (2 s)² - (x0 + v0 t + 1/2 a (1 s)²) = 14 m

x0 + v0 t + 2 a s² - x0 -v0 t - 1/2 a s² = 14 m

2 a s² - 1/2 a s² = 14 m

3/2 a s² = 14 m  

a = 14 m / (3/2 s²) = 9.3 m/s²