Hello!
I'm studying for physics and I am stuck on this question (I have the answer):
Two identical cars are travelling on a horizontal road, one of them is going at twice the speed of the other one. Both break using the same force. The slower car breaks completely after 20m. At what distance did the faster car come to a full stop? The answer was 80m. I just don't know how to arrive to it. Can anyone help? Thank you!
Newton's second law:
F=ma
if F and m are both identical, the a would be the same, where a=deceleration in m/s².
Using the kinematic equations for the first car,
vf²=vi²+2aΔx
vf=final velocity=0
vi=initial velocity
Δx=stopping distance
and solve for Δx.
Δx = -vi²/2a = 20 m
where a<0 for deceleration.
For the second car, vi is doubled, so
-(2vi&sp2;)/2a
=-4vi²/2a
=80 m
I saw this earlier and I tried using kinematic equations, but I could't reason 80m as the answer. When you plug in a value for a that is less then or greater then 0, you do not calculate 80m, at least I didn't. So, I am not sure what MathMate did, and I do not think that it is correct.
Use the following relationship, which is based on the conservation of mechanical energy:
Work=Kinetic Energy
F*d=1/2mv^2
Where
F=force
d=displacement
v=velocity
and
m=mass
The cars are identical, so F is the same for both cars. Solve for F:
F*d=1/2mv^2
F=(1/2mv^2)/d
For the first car, the car that stops a distance of 20m, the equation becomes the following:
F*20m=1/2mv^2
Solving for F:
F=(1/2m^2)/20m
*** The question states that the cars experience the same force and are identical.
Since F for both are cars are the same, set the equations equal to each other.
(1/2m^2)/d=(1/2m^2)/20m
Masses are equal for both cars and cancel out since the cars are identical.
(1/2v^2)/d=(1/2v^2)/20m
Further simplification gives the following:
v^2/d=v^2/20m
*** Remember, the second car has a velocity twice that of the first car
Let v1= velocity of the first car and v2= velocity of the second car.
v2=2v1, so the equation becomes the following:
[(2v)^2/]d=v^2/20m
4v^2/d=v^2/20m
Taking the inverse of the equation, the equation becomes the following:
d/4v^2=20m/v^2
Solving for d:
d=20m*(4v^2/v^2)
d=20m*4
d=80m
I hope this explanation wasn't too late, and best and good luck.
thank you so much!
Of course! I'll be happy to help you understand how to arrive at the answer.
To solve this problem, we can start by using the concept of braking distance, which is the distance a car travels after the brakes are applied until it comes to a complete stop. The braking distance depends on the initial speed of the car and the braking force applied.
In this case, we are given that the slower car comes to a complete stop after traveling a distance of 20m. We can represent the speed of the slower car as "v" and the braking distance as "d".
We can assume that the braking force is the same for both cars. Therefore, the braking distance is directly proportional to the initial speed. This means that the faster car, which is traveling at twice the speed of the slower car, will have a braking distance that is also twice as long.
So if the slower car comes to a stop after 20m, we can conclude that the faster car will come to a stop after traveling a distance of 2 times 20m, which is 40m.
However, the question asks for the distance at which the faster car comes to a full stop, not just when its brakes are applied. We know that the braking distance of the faster car is 40m, but it will also continue to move for a certain distance before coming to a full stop.
To find this additional distance, we need to consider the time it takes for the faster car to come to a stop. Since the braking force and mass are the same for both cars, the time it takes to stop will also be the same. This means that the faster car will cover the additional distance during the same amount of time it took for the slower car to come to a stop.
Since distance equals speed multiplied by time (d = v * t), and the speeds of both cars are proportional to their braking distances, we can also say that the additional distance covered by the faster car is proportional to the additional time it takes to stop.
Since the faster car needs to cover an additional 40m, which is twice the braking distance of the slower car, it will also take twice the additional time. Therefore, the additional distance covered by the faster car is also 2 times 20m, which is 40m.
Now, to find the total distance covered by the faster car, we add the braking distance to the additional distance. The braking distance is 40m, and the additional distance is 40m, so the total distance covered by the faster car before coming to a full stop is 40m + 40m = 80m.
Therefore, the answer to the question is 80m. I hope this explanation helps you understand how to arrive at the answer. Let me know if there's anything else I can assist you with!