Let your car have a mass of 825 kg. Your car is moving at a speed of 3.08 m/s when it strikes a parked car and stops in 0.21 s. What force did the parked car exert on it to stop it that quickly?
Vf = Vo + a*t = 0.
3.08 + a*0.21 = 0.
0.21a = -3.08.
a = -14.7 m/s^2.
F = M*a.
To find the force exerted by the parked car on your car, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
1. Calculate the acceleration of your car:
- We know the initial velocity (3.08 m/s) and the final velocity (0 m/s) when your car stops.
- Using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken, we can rearrange the formula to solve for acceleration:
a = (v - u) / t
a = (0 m/s - 3.08 m/s) / 0.21 s
a = -3.08 m/s / 0.21 s
a = -14.67 m/s^2
Note: The negative sign indicates that the acceleration is in the opposite direction of the initial velocity.
2. Use Newton's second law to calculate the force exerted by the parked car:
- The mass of your car is given as 825 kg.
- The acceleration of your car is -14.67 m/s^2 (negative due to the direction).
- The force can be calculated using the formula F = m * a:
F = 825 kg * (-14.67 m/s^2)
F = -12,102.75 N
Note: The negative sign indicates that the force exerted by the parked car is in the opposite direction of the initial velocity.
Therefore, the parked car exerted a force of approximately 12,102.75 N on your car to stop it that quickly.