A car is brought to rest in additions of 484 m using a constant acceleration of -8.0 m/s squared. What was the velocity of the car when the acceleration first began?
"additions" ? You mean "a distance" ?
v^2 = 2as = 2*8*484 = 7744
v = 88 m/s
To find the initial velocity of the car, we can use the equation:
vf^2 = vi^2 + 2ad
Where:
vf = final velocity (0 m/s, as the car is brought to rest)
vi = initial velocity (what we want to find)
a = acceleration (-8.0 m/s^2)
d = distance (484 m)
Plugging in the values:
0^2 = vi^2 + 2*(-8.0 m/s^2)*484 m
Simplifying the equation:
0 = vi^2 - 7744 m/s^2
Rearranging the equation and solving for vi:
vi^2 = 7744 m/s^2
vi = √7744 m/s^2
vi ≈ 88 m/s
Therefore, the velocity of the car when the acceleration first began was approximately 88 m/s.
To find the velocity of the car when the acceleration first began, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (which is 0 m/s, since the car comes to rest)
u = initial velocity (what we're trying to solve for)
a = acceleration (-8.0 m/s^2)
s = displacement (484 m)
Since the car comes to rest, the final velocity is 0. By substituting the known values into the equation, we can solve for the initial velocity (u):
0^2 = u^2 + 2*(-8.0 m/s^2)*484 m
Simplifying the equation:
0 = u^2 - 7744 m^2/s^2
Rearranging the equation:
u^2 = 7744 m^2/s^2
Taking the square root of both sides:
u = ±√(7744 m^2/s^2)
Therefore, the initial velocity of the car when the acceleration first began is u = ±88 m/s.