A common statistic in car tests is the standing (starting from rest) quarter-mile performance. A modern sports car can achieve a terminal speed (speed at the end of the quarter-mile) of 120 mph (193 km/h). How does the average acceleration compare to g

a = 0.36g

Hence, the acceleration of the car is 0.36 times the value of g

Explanation:

First we need to find the acceleration of this car. For that purpose we will use 3rdequation of motion:

2as = Vf² - Vi²

where,

a = acceleration = ?

s = distance = 0.25 miles =  (0.25 miles)(1609.34 m/1 mile) = 402.336 m

Vf = Terminal Velocity = (193 km/h)(1 h/3600 s)(1000 m/1 km) = 53.61 m/s

Vi = Initial Velocity = 0 m/s

Therefore,

2(a)(402.336 m) = (53.61 m/s)² - (0 m/s)²

a = (2874.15 m²/s²)/(2)(402.336 m)

a = 3.57 m/s²

Now, we compare it with value of g:

a/g = (3.57 m/s²)/(9.8 m/s²)

a = 0.36g

Hence, the acceleration of the car is 0.36 times the value of g

we need answer choices

the answer is b

explanation: