Question

The acceleration of automobiles is often given in terms of the time it takes to go from 0 mi/h to 60 mi/h. One of the fastest street legal cars, the Bugatti Veyron Super Sport, goes from 0 to 60 in 2.70 s. To the nearest integer, what is its acceleration in m/s^2? Please show work.

Answer 1

Answer:

9.934 m/s²

Explanation:

Given:

Initial speed of the Bugatti Veyron Super Sport = 0 mi/h

Final speed of the Bugatti Veyron Super Sport = 60 mi/h

Now,

1 mi/h = 0.44704 m / s

thus,

60 mi/h = 0.44704 × 60 = 26.8224 m/s

Time = 2.70 m/s

Now,

The acceleration (a) is given as:

$a=\frac{\textup{Change in speed}}{\textup{Time}}$

thus,

$a=\frac{26.8224 - 0 }{2.70}$

or

a = 9.934 m/s²