A car which is traveling at a velocity of 9.6 m/s undergoes an acceleration of 4.2 m/s2 over a distance of 450 m. How fast is it going after that acceleration?
d = Vi t + (1/2) a t^2
450 = 9.6 t + 2.1 t^2
solve quadratic for t
then
v = Vi + a t
v = 9.6 + 4.2 t
To find the final velocity of the car after the acceleration, we can use the equation:
vf² = vi² + 2 * a * d
Where:
- vf is the final velocity
- vi is the initial velocity
- a is the acceleration
- d is the distance
Given:
- vi = 9.6 m/s (initial velocity)
- a = 4.2 m/s² (acceleration)
- d = 450 m (distance)
We can substitute the values into the equation and solve for vf.
vf² = (9.6 m/s)² + 2 * (4.2 m/s²) * (450 m)
First, simplify the equation inside the parentheses:
vf² = 92.16 m²/s² + 2 * (4.2 m/s²) * (450 m)
Next, perform the multiplication within the parentheses:
vf² = 92.16 m²/s² + 3780 m²/s²
Now, combine the two terms:
vf² = 92.16 m²/s² + 3780 m²/s²
vf² = 3872.16 m²/s²
Finally, take the square root of both sides of the equation to solve for vf:
vf = √3872.16 m²/s²
vf ≈ 62.22 m/s
Therefore, the car will be going approximately 62.22 m/s after the acceleration.