Final answer:
The car accelerated for approximately 3.098 seconds.
Step-by-step explanation:
In this problem, we have a car that starts at an initial velocity of 12 m/s and accelerates uniformly until it reaches a final velocity of 19 m/s. The car travels a distance of 48 m while accelerating. We can use the equation:
vf = vi + at
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.
Plugging in the given values, we have:
19 m/s = 12 m/s + a * t
Subtracting 12 m/s from both sides, we get:
7 m/s = a * t
Dividing both sides by a, we get:
t = 7 m/s / a
Since the car travels a distance of 48 m while accelerating, we can use the equation:
vf^2 = vi^2 + 2a * d
where d is the distance.
Plugging in the given values, we have:
19^2 = 12^2 + 2a * 48
Simplifying, we get:
361 = 144 + 96a
Subtracting 144 from both sides, we get:
217 = 96a
Dividing both sides by 96, we get:
a = 217 / 96
Now we can substitute the value of a into the first equation to find the time:
t = 7 m/s / (217 / 96)
Simplifying, we get:
t = 7 * 96 / 217
Calculating, we find that t ≈ 3.098 seconds.