How far does the car travel before stopping?

[collegegirl]

Homework Statement

Car: mass = 200g
Moving at 30 m/s
Braking force of 10000N.
How far does the car travel before stopping?

Homework Equations

W=m x N
w = F (delta x)

The Attempt at a Solution

I found work, (2000)(10000) = 20,000,000
Then rearranged the second equation to get delta x, and got 2000 as my answer. The answer is 90m and I'm not getting it. Any suggestions?

 

[asleight]

Homework Statement

Car: mass = 200g
Moving at 30 m/s
Braking force of 10000N.
How far does the car travel before stopping?

You can go one of two routes:

(1) $$\vec{F}=m\vec{a}$$ and $$\vec{v}_f^2=\vec{v}_i^2+2\vec{a}\Delta x$$, or

(2) $$\Delta E_k=\vec{W}$$, where $$\vec{W}=\vec{F}\Delta x$$.

 

[thomate1]

I would like to clarify a few things about this problem. First, the mass of the car is given in grams, which is an unusual unit for a car's mass. It would be more appropriate to convert it to kilograms (200g = 0.2kg). Additionally, the units for velocity (m/s) and force (N) are both in SI units, which is good.

Secondly, the work-energy theorem (W = F * d) is not applicable in this situation because it assumes a constant force. In this problem, the braking force is not constant, as it decreases as the car slows down. Therefore, we cannot use this equation to find the distance traveled.

To solve this problem, we need to use the equations of motion, specifically the one that relates distance, velocity, acceleration, and time (d = v0t + 1/2at^2). In this case, the initial velocity (v0) is 30 m/s, the acceleration (a) is -10000N/0.2kg = -50000 m/s^2 (negative because it is in the opposite direction of motion), and we are trying to find the distance (d) traveled before the car stops (v = 0 m/s).

Plugging these values into the equation, we get:

0 = (30 m/s)t + 1/2(-50000 m/s^2)t^2

Simplifying, we get:

0 = 30t - 25000t^2

Solving for t using the quadratic formula, we get t = 0.0012 seconds.

Finally, we can plug this value back into the original equation to find the distance traveled:

d = (30 m/s)(0.0012 seconds) + 1/2(-50000 m/s^2)(0.0012 seconds)^2 = 0.036 m = 3.6 cm

Therefore, the car travels approximately 3.6 cm before coming to a complete stop. This is a very short distance, highlighting the importance of safe braking techniques and maintaining a safe speed while driving.