Police find skid marks 65.0 m long on a highway showing where the car made an emergency stop. Assuming the acceleration was -10.3 m/s^ ( about the maximum for dry pavement) how fast was the car going?
To find the initial speed of the car, we can use the kinematic equation for motion:
v² = u² + 2aS
Where:
- v is the final velocity (0 m/s since the car made an emergency stop)
- u is the initial velocity (what we need to find)
- a is the acceleration (-10.3 m/s², in this case)
- S is the distance traveled (65.0 m)
Rearranging the equation to solve for u gives us:
u = √(v² - 2aS)
u = √(0 - 2(-10.3)(65.0))
Now, let's plug in the values and calculate:
u = √(0 - 2(-10.3)(65.0))
u = √(0 + 1339)
u = √1339
u ≈ 36.61 m/s
Therefore, the car was going approximately 36.61 m/s before making the emergency stop.
To find the initial velocity (speed) of the car, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, as the car came to a stop)
u = initial velocity (unknown)
a = acceleration (-10.3 m/s^2)
s = distance (65.0 m)
Rearranging the equation, we get:
u^2 = v^2 - 2as
Since the final velocity (v) is 0 m/s and the distance (s) is 65.0 m, we can substitute these values into the equation:
u^2 = 0^2 - 2(-10.3)(65.0)
Simplifying:
u^2 = 0 - (-1339)
u^2 = 1339
Taking the square root of both sides:
u = √1339
u ≈ 36.60 m/s
Therefore, the initial velocity (speed) of the car was approximately 36.60 m/s.
V^2 = Vo^2 + 2a*d.
V = 0.
a = -10.3 m/s^2.
d = 65 m.
Vo = ?.