A driver of an automobile traveling at a constant speed of 20meter per second suddenly applies a brake and the automobile comes to rest in 2.0 seconds after skidding for a certain distance. what is the length of the skid distance?
a = -20/2 = -10 m/s^2
s = 20t + 1/2 at^2
Well, if the automobile comes to a rest after applying the brakes, I hope it also remembered to send a thank-you note to the brakes for their assistance. As for the skid distance, we can calculate it using the formula:
Distance = Initial Velocity * Time + (1/2) * Acceleration * Time^2
But since we know the automobile comes to rest, the final velocity would be 0. And since the automobile is skidding, the acceleration should be negative. So, we can rewrite the formula as:
Distance = Initial Velocity * Time + (1/2) * Acceleration * Time^2
Distance = (20 m/s) * (2 s) + (1/2) * (-20 m/s^2) * (2 s)^2
After crunching those numbers, we find that the length of the skid distance would be -40 meters. But don't worry, it's just the magnitude of the distance that matters, so we can simply consider it as a positive 40 meters. Just make sure not to apply brakes like that on the dance floor. It won't end well.
To find the length of the skid distance, we need to use the equation of motion for constant acceleration:
\[ v^2 = u^2 + 2as \]
where:
- \(v\) is the final velocity (which is 0 since the automobile comes to rest)
- \(u\) is the initial velocity (which is 20 m/s)
- \(a\) is the acceleration (unknown)
- \(s\) is the distance traveled (unknown)
Since the automobile comes to rest, the final velocity \(v\) is 0. We can rearrange the equation of motion to solve for \(s\):
\[ 0 = (20 \ \text{m/s})^2 + 2a \cdot s \]
Simplifying:
\[ 0 = 400 + 2a \cdot s \]
Considering that the stopping time is 2.0 seconds, we know that the final velocity is \(v = 0\) and the initial velocity was \(u = 20 \ \text{m/s}\). We can use the equation of motion for uniform acceleration:
\[ v = u + at \]
Substituting the values into the equation, we have:
\[ 0 = 20 \ \text{m/s} + a (2.0 \ \text{s}) \]
Simplifying:
\[ -20 \ \text{m/s} = 2a \ \text{m/s}^2 \]
Dividing both sides by 2:
\[ -10 \ \text{m/s} = a \ \text{m/s}^2 \]
Now we have two equations:
\[ 0 = 400 + 2a \cdot s \]
\[ -10 \ \text{m/s} = a \ \text{m/s}^2 \]
We will solve these equations simultaneously to find the value of \(s\).
To find the length of the skid distance, we can use the equation of motion for uniformly accelerated motion:
\[v^2 = u^2 + 2as\]
Where:
- \(v\) is the final velocity (0 m/s, as the automobile comes to rest),
- \(u\) is the initial velocity (20 m/s),
- \(a\) is the acceleration (unknown),
- \(s\) is the distance (unknown).
Since the initial velocity is positive and the final velocity is zero (as the automobile comes to rest), we can rewrite the equation as:
\[0^2 = (20)^2 + 2a \cdot s\]
Simplifying the equation gives:
\[0 = 400 + 2a \cdot s\]
As the car is skidding, it experiences negative acceleration due to the braking, which means the acceleration \(a\) is negative. Therefore, we can rewrite the equation as:
\[0 = 400 - 2|a| \cdot s\]
Rearranging the equation to solve for the skid distance \(s\), we have:
\[s = \frac{400}{2|a|}\]
Now, we can plug in the value for time taken to stop (2.0 seconds) to solve for acceleration \(a\). When a car comes to rest, the final velocity is zero, so we can use the equation of motion:
\[v = u + at\]
\[0 = 20 + a \cdot 2.0\]
Solving for \(a\), we get:
\[a = \frac{-20}{2.0}\]
\[a = -10 \, \text{m/s}^2\]
Substituting this value of acceleration back into the equation for \(s\), we can find the skid distance:
\[s = \frac{400}{2 \cdot |-10|}\]
\[s = \frac{400}{20}\]
\[s = 20 \, \text{meters}\]
Therefore, the length of the skid distance is 20 meters.