A car has an initial velocity of 18 m/s and an acceleration of 9 m/s2. At what time and distance will the car has before it comes to a stop?
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s2. At what time and distance will the car has before it comes to a stop?
A car has an initial velocity of 18 m/s and an acceleration of 9 m/s2. At what time and distance will the car has before it comes to a stop?
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Answer:
Given:
- Initial velocity (u) = 18 m/s
- Acceleration (a) = -9 m/s^2 (negative since it's deceleration)
- Final velocity (v) = 0 m/s
- Distance or displacement (s) = ?
v^2 = u^2 + 2as
(0 m/s)^2 = (18 m/s)^2 + 2(-9 m/s^2)s
0 = 324 m^2/s^2 - 18 m/s^2 * 2s
Solve for s:
324 m^2/s^2 - 36 m/s^2s = 0
-36 m/s^2s = -324 m^2/s^2
s = -324 m^2/s^2 / -36 m/s^2
s = 9 m
To find the time it takes for the car to stop.
v = u + at
0 m/s = 18 m/s + (-9 m/s^2) * t
Simplifying:
-18 m/s = -9 m/s^2 * t
t = -18 m/s / -9 m/s^2
t = 2 s