A driver in a car traveling at a speed of 70 km/h sees a cat 129 m away on the road. How long will it take for the car to accelerate uniformly to a stop in exactly 125 m? answer in units of s
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To find the time it takes for the car to accelerate uniformly to a stop, we can use the kinematic equation:
\(v^2 = u^2 + 2aS\)
where:
- \(v\) is the final velocity (0 m/s since the car comes to a stop),
- \(u\) is the initial velocity,
- \(a\) is the acceleration,
- \(S\) is the displacement.
We know the initial velocity (\(u\)) is 70 km/h. We need to convert it to meters per second (m/s).
\(u = 70 \, \text{km/h} = \frac{{70 \times 1000}}{{3600}} \, \text{m/s} \approx 19.44 \, \text{m/s}\)
The displacement (\(S\)) is given as 125 m.
Now, rearranging the equation to solve for acceleration (\(a\)) before substituting the known values:
\(v^2 = u^2 + 2aS \)
\(0^2 = (19.44)^2 + 2a(125) \)
\(0 = 378.13 + 250a \)
\(-378.13 = 250a \)
\(a = \frac{{-378.13}}{{250}} \approx -1.51 \, \text{m/s}^2\)
Now, we can use the equation \(v = u + at\) to find the time (\(t\)) it takes for the car to decelerate to a stop. Again, we can substitute the known values:
\(v = u + at \)
\(0 = 19.44 + (-1.51)t \)
\(1.51t = 19.44 \)
\(t = \frac{{19.44}}{{1.51}} \approx 12.88 \, \text{s}\)
Therefore, it will take approximately 12.88 seconds for the car to accelerate uniformly to a stop in exactly 125 m.