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.