The driver in a car of total mass 800kg and which is travelling at 20m / s suddenly observes a stationary dog in his path 50m ahead . if the car brakes can exert a force of 2000N, what will likely happen

Well, it seems like this driver is in quite the "ruff" situation! Let's do some calculations to figure out what will likely happen.

To stop the car in time, the driver needs to apply enough braking force to overcome the car's initial momentum. The initial momentum of the car can be calculated using the formula p = mv, where p is momentum, m is mass, and v is velocity.

Given:
- Mass of the car (m) = 800 kg
- Velocity of the car (v) = 20 m/s

Initial momentum of the car = m × v = 800 kg × 20 m/s = 16,000 kg·m/s

Now, let's determine if the car can stop in time. To do this, we need to calculate the force required to bring the car to a stop in 50 meters. The formula to calculate force is F = Δp/Δt, where F is force, Δp is the change in momentum, and Δt is the time interval.

Given:
- Change in momentum (Δp) = -16,000 kg·m/s (since the car needs to come to a stop)
- Distance (Δx) = 50 m
- Initial velocity (v) = 20 m/s

Δp = m × (v_final - v_initial)
Δp = 800 kg × (0 - 20 m/s)
Δp = -16,000 kg·m/s

Now, let's calculate the time interval (Δt) using the distance and velocity:
Δx = v_avg × Δt (since the car is braking uniformly)
50 m = (0 m/s + 20 m/s)/2 × Δt
50 m = 10 m/s × Δt
Δt = 5 s

Finally, to find the force required:
F = Δp/Δt
F = -16,000 kg·m/s / 5 s
F = -3,200 N

The braking force required to bring the car to a stop in time is -3,200 N, which is greater than the car's maximum braking force of 2,000 N. Therefore, it is likely that the car will not be able to stop in time and, unfortunately, the doggo might get a "taste" of car-becue. Remember to drive safely and keep an eye out for any furry friends on the road!

To determine what will likely happen, we need to analyze the forces acting on the car.

1. The initial forward momentum of the car is given by:
Initial momentum (p_initial) = mass of the car x velocity of the car
= 800 kg x 20 m/s
= 16,000 kg·m/s

2. When the car applies brakes, it exerts a force of 2000 N opposing the forward motion.

3. To bring the car to a stop, the force exerted by the brakes (2000 N) needs to overcome the forward momentum of the car (16,000 kg·m/s) and also reduce the car's kinetic energy.

4. The deceleration (negative acceleration) of the car can be calculated using Newton's second law:
Force = mass x acceleration
2000 N = 800 kg x acceleration
Acceleration = 2000 N / 800 kg
Acceleration = 2.5 m/s^2

5. The deceleration will cause the car to slow down until it comes to a stop. However, whether the car will stop in time to avoid hitting the dog will depend on the specific circumstances such as the distance between the car and the dog, the reaction time of the driver, and the grip of the tires on the road.

In conclusion, if the car can decelerate with an acceleration of 2.5 m/s^2, it will likely be able to stop in time to avoid hitting the dog.

To determine what will likely happen when the driver of the car observes a stationary dog in their path, we need to consider the forces acting on the car.

First, let's calculate the initial kinetic energy of the car. The kinetic energy (KE) is given by the equation KE = 1/2 * mass * velocity^2.

Mass of the car (m) = 800 kg
Velocity of the car (v) = 20 m/s

Initial kinetic energy (KE_initial) = 1/2 * 800 kg * (20 m/s)^2
= 1/2 * 800 kg * 400 m^2/s^2
= 320,000 J

Next, let's evaluate the force required to stop the car. The force (F) required to stop an object depends on its mass and the acceleration (a) applied. Using Newton's second law of motion (F = ma), if the car is at rest, the force required to stop it will be the product of its mass and the deceleration.

Mass of the car (m) = 800 kg
Deceleration (a) = (0 m/s - 20 m/s) / (0 s - t) = -20 m/s^2 (negative sign because it's deceleration)

Force required to stop the car (F) = mass (m) * deceleration (a)
= 800 kg * (-20 m/s^2)
= -16,000 N

Since the car's brakes can exert a force of 2,000 N, which is less than the force required to stop the car (-16,000 N), the car will not be able to come to a complete stop in time to avoid hitting the dog.

Therefore, it is likely that the car will hit the dog. The extent of the impact will depend on variables such as the car's speed, how hard the brakes are applied, and the dog's size and position relative to the car.

f = m a ... 2000 = 800 * a ... a = 2.5 m/s^2

it takes 8 s to stop ... 20 m/s / 2.5 m/s^2

average velocity during stop
... (20 + 0) / 2 = 10 m/s

... sorry, pooch ...