A car initially going 30ft/sec brakes at a constant rate (constant negative acceleration) coming to a stop in 4 seconds. How far did the car travel?
simple minded way: average velocity was 15ft/sec, for 4 seconds, 60ft
Nerd way:
distance= Vi*t- 1/2 a t^2
where a= changevelociy/time= 30/4
distance= 30*4-1/2 30/4 * 4^2 and that works out to ...duh...60 ft.
To find the distance the car traveled, we can use the kinematic equation:
\[d = v_0t + \frac{1}{2}at^2\]
Where:
- \[d\] is the distance traveled by the car
- \(v_0\) is the initial velocity, which is 30 ft/sec
- \(t\) is the time taken to stop, which is 4 seconds
- \(a\) is the acceleration, which is the negative of the braking rate
Since the car comes to a stop, its final velocity is 0 ft/sec. The final velocity can be calculated using the equation:
\[v = v_0 + at\]
As the car comes to a stop, \(v = 0\), so we can solve for \(a\):
\[0 = 30 + a(4)\]
Rearranging the equation, we have:
\[120 = 4a\]
Solving for \(a\):
\[a = \frac{120}{4} = 30\,ft/sec^2\]
Now that we have the value of \(a\), we can substitute it into the initial equation to find the distance traveled:
\[d = (30)(4) + \frac{1}{2}(-30)(4)^2\]
Simplifying:
\[d = 120 - 240\]
\[d = -120\,ft\]
The negative sign indicates that the car traveled in the opposite direction of its initial velocity. So, the car traveled 120 feet in the opposite direction.