a car accelerates uniformly from rest to a speed of 21.8 km/h in 5.9 seconds.find the distance it travels during this time. answer in units of meters
To find the distance traveled by the car, we will use the constant acceleration equation:
vf = vi + at
Where:
vf = final velocity
vi = initial velocity (which is 0 since the car starts from rest)
a = acceleration
t = time
We need to convert the final velocity from km/h to m/s since the time is given in seconds. To do this, we will use the conversion factor: 1 km/h = 0.2778 m/s.
vf = 21.8 km/h = 21.8 × 0.2778 m/s
vf = 6.05 m/s (approximately)
Now, we can rewrite the equation as:
6.05 m/s = 0 + a × 5.9 s
Simplifying the equation:
6.05 m/s = 5.9 s × a
Now, we can isolate the acceleration:
a = 6.05 m/s ÷ 5.9 s
a ≈ 1.02 m/s² (rounded to two decimal places)
Now that we have the acceleration, we can use the kinematic equation:
d = vit + 0.5at²
Where d is the distance, vi is the initial velocity (0 m/s), a is the acceleration (1.02 m/s²), and t is the time (5.9 s).
Substituting the values into the equation:
d = 0 × 5.9 s + 0.5 × 1.02 m/s² × (5.9 s)²
Simplifying and calculating:
d ≈ 0 + 0.5 × 1.02 m/s² × 34.81 s²
d ≈ 0.5 × 1.02 m/s² × 34.81 s²
d ≈ 0.51 m/s² × 34.81 s²
d ≈ 17.7671 m²/s² (approximately)
The distance traveled by the car during this time is approximately 17.77 meters.