A car is traveling at 30 m/s. It slows down with an acceleration of -4 m/s^2. How far does the car travel before it stops?
To find the distance the car travels before it stops, we can use the equation of motion:
(vf)^2 = (vi)^2 + 2aΔx
Where:
- vf is the final velocity, which is 0 m/s since the car stops
- vi is the initial velocity, which is 30 m/s
- a is the acceleration, which is -4 m/s^2
- Δx is the distance traveled
Substituting the known values into the equation, we can solve for Δx:
(0)^2 = (30)^2 + 2(-4)Δx
0 = 900 - 8Δx
8Δx = 900
Δx = 900 / 8
Δx ≈ 112.5 meters
Therefore, the car travels approximately 112.5 meters before it stops.
To find the distance the car travels before it stops, we can use the kinematic equation:
\(v_f^2 = v_i^2 + 2a \Delta x\)
Where:
\(v_f\) is the final velocity (which is 0 m/s since the car stops),
\(v_i\) is the initial velocity (30 m/s),
\(a\) is the acceleration (-4 m/s^2), and
\(\Delta x\) is the distance traveled.
Rearranging the equation, we get:
\(\Delta x = \frac{{v_f^2 - v_i^2}}{{2a}}\)
Plugging in the known values, we have:
\(\Delta x = \frac{{0^2 - 30^2}}{{2(-4)}}\)
\(\Delta x = \frac{{0 - 900}}{{-8}}\)
\(\Delta x = \frac{{900}}{{8}}\)
\(\Delta x = 112.5\) meters
Therefore, the car travels 112.5 meters before it stops.
t = 30 / 4 ... s
average velocity = (30 + 0) / 2 ... m/s
stopping distance = (average velocity * t