What average force is required to stop an 1900 car in 6.0 if the car is traveling at 80 ?
kilograms? pounds? , seconds? miles per hour, kilometers per hour ?
F = m delta V / delta t
= 1900 (0-80)/6
in whatever consistent units
I can not give you a proper answer without the units being stated.
To find the average force required to stop a car, we need to use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration. In this case, the car's mass is given as 1900 kg, and we need to find the acceleration to calculate the force.
First, we need to find the deceleration (negative acceleration) of the car. We can use the kinematic equation:
v^2 = u^2 + 2as,
where:
v = final velocity (0 m/s, since the car comes to a stop)
u = initial velocity (80 m/s)
a = acceleration (deceleration in this case)
s = distance traveled (6.0 m)
Rearranging the equation, we get:
a = (v^2 - u^2) / (2s)
Substituting the values, we have:
a = (0^2 - 80^2) / (2 * 6.0)
Simplifying:
a = (-6400) / 12
a = -533.33 m/s^2 (deceleration)
Now, we can use Newton's second law to calculate the average force:
F = m * a
Substituting the values, we have:
F = 1900 * (-533.33)
Calculating:
F ≈ -1,013,993.33 N
Given that force is a vector quantity, the negative sign indicates that the force is in the opposite direction of motion (opposite to the initial velocity).
Therefore, the average force required to stop the car is approximately 1,013,993.33 Newtons (N).