PHYSICS HELP

Question

Calculate the force needed to bring a 1070–kg car to rest from a speed of 89.0 km/h in a distance of 110 m (a fairly typical distance for a non-panic stop).

 

Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.34 m. Calculate the force exerted on the car and compare it with the force found in part (a).

Answer 1

• F = ma
• We must first calculate a, the acceleration, before applying the F= ma equation to determine the force.
• In order to calculate the acceleration, we must first use the following equation of kinematics, Vf² = Vi² + 2ax.
• V_f = final velocity = 0 meters/second. (The vehicle is stopped)
• V_i = initial velocity = 89 kilometers/hour, however, this must be converted to meters/second = 24.72 m/s
• a = acceleration = quantity that we are attempting to calculate
• x = distance = 110 meters
• rearranging the formula, a = (V_f² - V_i²)/2X = (0 - 611.2)/2(110) ≅ 2.78 m/s² (In Magnitude)
  
• F = ma = (1070 Kg)(2.78 m/s²) ≅ 2972 Kg·m/s² or 2972 Newtons ( 1 Newton = 1 Kg·m/s²)
• For the second part, just replace X=110 to X=2.34
• a = (Vf² - Vi²)/2X = (0 - 611.2)/2(2.34) ≅ 130.6 m/s² 
• F = ma = (1070 Kg)(130.6 m/s²) ≅ 139,740 Newtons
• This force is about 47 times greater than the force in the initial portion
• The same mass is brought to a halt from the same speed; however, this occurs over a much shorter distance and time interval in the second part. Notice that the Force ratio is equal to the ratio of Accelerations