Final answer:
To bring a car to rest from a speed of 85.0 km/h in a distance of 110 m, the force required is -2466 N. If the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m, the force exerted on the car is -21420 N. The force exerted during the collision is much greater than the force required for the non-panic stop.
Step-by-step explanation:
(a) Calculating the force needed to bring the car to rest:
First, we need to convert the speed from km/h to m/s. Since 1 km/h is equal to 0.28 m/s, the speed becomes 85.0 km/h x 0.28 m/s = 23.8 m/s.
Next, we calculate the deceleration of the car using the formula a = (v^2 - u^2) / (2d), where v is the final velocity (0 m/s), u is the initial velocity (23.8 m/s), and d is the stopping distance (110 m). Plugging in the values, we get a = (0 - 23.8^2) / (2 x 110) = -2.74 m/s^2 (negative because it's deceleration).
Finally, we can calculate the force using the formula F = ma, where m is the mass of the car (900 kg) and a is the deceleration. Plugging in the values, we get F = 900 kg x (-2.74 m/s^2) = -2466 N (negative because it's acting in the opposite direction).
(b) Comparing the force with a collision at full speed:
Since the car stops in a much shorter distance (2.00 m) compared to the non-panic stop (110 m), the force exerted on the car during the collision will be much greater. To calculate the force, we use the formula F = ma, where m is the mass of the car (900 kg) and a is the deceleration. Plugging in the values, we get F = 900 kg x (-23.8 m/s^2) = -21420 N (negative because it's acting in the opposite direction).
The force exerted during the collision is much greater than the force required for the non-panic stop, indicating a much greater impact.