What average force is required to stop an 1100 kg car in 9.1 s if it is traveling at 83 km/h?

12

434

2787 Newtons

To find the average force required to stop a car, we need to use Newton's second law of motion, which states that the force (F) acting on an object is equal to the mass (m) of the object multiplied by the acceleration (a) it experiences. In this case, the car's initial velocity (v1) and final velocity (v2) can be used to calculate the acceleration.

First, let's convert the car's initial velocity from km/h to m/s. We can do this by dividing the given value (83 km/h) by 3.6 since 1 km/h is equivalent to 1/3.6 m/s.

v1 = 83 km/h ÷ 3.6 = 23.06 m/s

Next, we need to determine the change in velocity. Since the car is coming to a stop, its final velocity (v2) will be 0 m/s. Therefore, the change in velocity (Δv) can be calculated as:

Δv = v2 - v1 = 0 - 23.06 m/s = -23.06 m/s

Now, we can calculate the acceleration (a) using the formula:

a = Δv / t

where Δv is the change in velocity and t is the time taken to come to a stop. In this case, the given time is 9.1 seconds.

a = -23.06 m/s / 9.1 s ≈ -2.53 m/s²

Since the car is decelerating (negative acceleration), the force required to stop it will be in the opposite direction of its motion. Therefore, we take the absolute value of the acceleration.

|a| = 2.53 m/s²

Finally, we can calculate the average force (F) by using Newton's second law of motion:

F = m * |a|

where m is the mass of the car, given as 1100 kg.

F = 1100 kg * 2.53 m/s² ≈ 2783 N

Therefore, the average force required to stop the car is approximately 2783 Newtons.