Answer:
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
