A car is traveling at Vx =20 m/s. The driver applies the brakes, and the car slows with ax = -4m/s^2. What is the stopping distance
To find the stopping distance of the car, we can use the formula:
stopping distance = (initial velocity^2) / (2 * acceleration)
Given:
Initial velocity, Vx = 20 m/s
Acceleration, ax = -4 m/s^2
Substituting the given values into the formula, we have:
stopping distance = (20^2) / (2 * -4)
To simplify the equation, we can solve the numerator first.
20^2 = 400
Now, let's simplify the denominator:
2 * -4 = -8
We can now substitute the simplified values into the equation:
stopping distance = 400 / -8
When dividing a positive number by a negative number, the result is negative:
stopping distance = -50
Therefore, the stopping distance of the car is 50 meters.
To find the stopping distance of the car, we can use the following equation:
\(v_f^2 = v_i^2 + 2a_d\)
Where:
\(v_f\) is the final velocity of the car (which is 0, as the car stops),
\(v_i\) is the initial velocity of the car (given as \(V_x = 20 \, m/s\)), and
\(a\) is the acceleration of the car (given as \(a_x = -4 \, m/s^2\)).
Plugging in the values, we have:
\(0 = (20 \, m/s)^2 + 2(-4 \, m/s^2)d\)
Simplifying the equation:
\(0 = 400 \, m^2/s^2 - 8d\)
Rearranging the equation:
\(8d = 400 \, m^2/s^2\)
\(d = \frac{400 \, m^2/s^2}{8}\)
\(d = 50 \, m\)
Thus, the stopping distance of the car is 50 meters.