To find the acceleration and time to stop, we can use the following equations of motion:
- Final velocity squared (v^2) = Initial velocity squared (u^2) + 2 * acceleration (a) * distance (s)
- Final velocity (v) = Initial velocity (u) + acceleration (a) * time (t)
- Distance (s) = Initial velocity (u) * time (t) + (1/2) * acceleration (a) * time squared (t^2)
Given: Initial velocity (u) = 30 m/s Distance (s) = 100 m
We need to find the acceleration (a) and the time to stop (t).
Using equation (1):
0 = 30^2 + 2 * a * 100
900 = 200a
a = 900 / 200 a = 4.5 m/s²
Therefore, the acceleration is 4.5 m/s².
To find the time to stop, we can use equation (3):
100 = 30t + (1/2) * 4.5 * t^2
Simplifying the equation:
4.5t^2 + 30t - 100 = 0
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
a = 4.5, b = 30, c = -100
t = (-30 ± √(30^2 - 4 * 4.5 * -100)) / (2 * 4.5)
t = (-30 ± √(900 + 1800)) / 9
t = (-30 ± √2700) / 9
Since time cannot be negative in this context, we take the positive root:
t = (-30 + √2700) / 9
t ≈ 2.92 seconds
Therefore, the time taken for the car to stop is approximately 2.92 seconds.