Question

A sports car accelerates from rest to 27.8 m/s in 6.2 s.

a. Calculate its acceleration in m/s² (2 pts)
b. How many g’s is this (1 g = 9.8 m/s²) (1 pt)

Option 1: a. 4.48 m/s², b. 0.458 g
Option 2: a. 3.74 m/s², b. 0.381 g
Option 3: a. 5.09 m/s², b. 0.520 g
Option 4: a. 6.62 m/s², b. 0.676 g

Answer 1

The acceleration of the sports car is 4.48 m/s² and this equates to approximately 0.458 g's when compared to the acceleration due to gravity. The correct answer is Option 1 a. 4.48 m/s², b. 0.458 g.

Step-by-step explanation:

To calculate the acceleration of the sports car, we use the formula a = Δv / Δt, where Δv is the change in velocity and Δt is the change in time. The sports car accelerates from rest (initial velocity, vi = 0) to a final velocity (vf = 27.8 m/s) over a period of time (t = 6.2 s).

Using the formula:

a = (vf - vi) / t = (27.8 m/s - 0) / 6.2 s = 4.48 m/s²

The acceleration due to gravity (g) is 9.8 m/s², so to find the number of g's, we divide the car's acceleration by the acceleration due to gravity:

Number of g's = a / g = 4.48 m/s² / 9.8 m/s² = 0.457 g (rounded to three decimal places)

Therefore, the correct answer is Option 1: a. 4.48 m/s², b. 0.458 g (rounded to three decimal places).