a 1240 kg automobile stops in a distance of 41m. if initial velocity was 22.5m/s, what force did the road exert on the car to stop it? Answer pls... Thank You

how long does it take to stop, with acceleration a?

22.5t + 1/2 at^2 = 41
Now, with that t,
22.5 + at = 0
That gives you the value of a.

F = ma

How to get acceleration? and time?

To calculate the force exerted by the road on the car, we can use the equation for force:

force = mass × acceleration

First, let's find the acceleration of the car. We can use the formula:

acceleration = (final velocity - initial velocity) / time

Since the car stops, the final velocity is 0. Therefore, the equation becomes:

acceleration = (0 - 22.5 m/s) / (unknown time)

To find the time it takes for the car to stop, we can use the kinematic equation:

distance = (initial velocity × time) + (0.5 × acceleration × time^2)

We know the distance the car stopped (41 m) and the initial velocity (22.5 m/s), but we need the time to solve for the acceleration. Rearranging the equation, we get:

0 = (0.5 × acceleration × time^2) + (initial velocity × time) - distance

This is a quadratic equation, and we can solve for time using the quadratic formula:

time = (-b ± √(b^2 - 4ac)) / 2a

Now, substitute the values into the equation:

a = 0.5 × acceleration
b = initial velocity
c = -distance

Solving this equation gives us two possible values for time, but since it takes a positive amount of time for the car to stop, we choose the positive root.

With the time known, we can go back to the equation for acceleration:

acceleration = (final velocity - initial velocity) / time

Plug in the values:

acceleration = (0 - 22.5 m/s) / (calculated time)

Now we can calculate the force exerted by the road on the car:

force = mass × acceleration

force = (1240 kg) × (calculated acceleration)

Evaluating this expression will give us the force exerted by the road on the car.