A car moving initially at a speed of 50 mi/h (=80 km/h) and weighing 3000 lb (=13,000 N) isbrought to a stop in a distance of 200 ft (=61 m). Find (a) the braking force and (b) the timerequired to stop. Assuming the same braking force, find (c) the distance and (d ) the timerequired to stop if the car were going 25 mi/h (=40 km/h) initially.

To solve this problem, we'll use the equations of motion and the relationship between force, mass, and acceleration.

(a) The braking force

First, we need to convert the initial speed and stopping distance into consistent units. Let's use the metric system. The initial speed is 80 km/h, which is approximately 22.22 m/s (since 1 km/h = 0.27778 m/s). The stopping distance is 61 m.

We can use the equation v^2 = u^2 + 2as to find the deceleration (a), where v is the final speed, u is the initial speed, s is the distance, and a is the acceleration (which will be negative since it's deceleration).

Here, v = 0 (since the car comes to a stop), u = 22.22 m/s, and s = 61 m. Plugging these values into the equation gives us:

0 = (22.22 m/s)^2 + 2a61 m => a = -(22.22 m/s)^2 / (2*61 m) = -8.13 m/s^2

The force required to stop the car can be found using the equation F = ma, where m is the mass and a is the acceleration. The mass of the car is 1300 kg (since 1 N = 1 kg*m/s^2). So, the force is:

F = 1300 kg * 8.13 m/s^2 = 10569 N

(b) The time required to stop

We can use the equation v = u + at to find the time (t), where v is the final speed, u is the initial speed, a is the acceleration, and t is the time. Here, v = 0, u = 22.22 m/s, and a = -8.13 m/s^2. Plugging these values into the equation gives us:

0 = 22.22 m/s - 8.13 m/s^2 * t => t = 22.22 m/s / 8.13 m/s^2 = 2.73 s

(c) The distance required to stop if the car were going 40 km/h initially

If the car were going 40 km/h initially, that's approximately 11.11 m/s. We can use the same deceleration as before (since the problem states to assume the same braking force). So, we can use the equation v^2 = u^2 + 2as to find the new stopping distance (s). Here, v = 0, u = 11.11 m/s, and a = -8.13 m/s^2. Plugging these values into the equation gives us:

0 = (11.11 m/s)^2 + 2*(-8.13 m/s^2)s => s = (11.11 m/s)^2 / (28.13 m/s^2) = 7.6 m

(d) The time required to stop if the car were going 40 km/h initially

We can use the equation v = u + at to find the new time (t). Here, v = 0, u = 11.11 m/s, and a = -8.13 m/s^2. Plugging these values into the equation gives us:

0 = 11.11 m/s - 8.13 m/s^2 * t => t = 11.11 m/s / 8.13 m/s^2 = 1.37 s

So, the braking force is approximately 10569 N, the time required to stop from 80 km/h is approximately 2.73 s, the distance required to stop from 40 km/h is approximately 7.6 m, and the time required to stop from 40 km/h is approximately 1.37 s.