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

seeing the graph would

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day