A driver in a car traveling at a speed of 67 km/h sees a cat 104 m away on the road.
How long will it take for the car to accelerate uniformly to a stop in exactly 102 m?
To determine how long it will take for the car to accelerate uniformly to a stop in exactly 102 m, we can use the equations of motion.
The first step is to convert the given speed from km/h to m/s. We know that 1 km/h is equal to 1000 m/3600 s. So, the speed of the car in m/s is:
67 km/h * (1000 m/1 km) * (1 h/3600 s) = 18.61 m/s (rounded to 2 decimal places)
Next, we can use the equation of motion:
v^2 = u^2 + 2as
where v is the final velocity (0 m/s since the car is coming to a stop), u is the initial velocity (18.61 m/s), a is the acceleration, and s is the distance traveled.
Rearranging the equation to solve for acceleration (a), we have:
a = (v^2 - u^2) / (2s)
Plugging in the values we know:
a = (0 m/s)^2 - (18.61 m/s)^2) / (2 * 102 m) = -344.24 m^2/s^2 (rounded to 2 decimal places)
The negative sign indicates that the car is decelerating.
Now, we can use the formula of uniform acceleration to find the time taken (t) to cover the distance (s) with the given acceleration (a):
s = ut + 0.5at^2
Plugging in the values:
102 m = (18.61 m/s) * t + 0.5 * (-344.24 m^2/s^2) * t^2
Rearranging the equation and converting it into a quadratic equation:
0.5 * (-344.24 m^2/s^2) * t^2 + (18.61 m/s) * t - 102 m = 0
We can solve this quadratic equation for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
where a = 0.5 * (-344.24 m^2/s^2), b = 18.61 m/s, and c = -102 m.
Solving for t, we get:
t = (-18.61 ± √((18.61)^2 - 4 * 0.5 * (-344.24) * (-102))) / (2 * 0.5 * (-344.24))
This gives us two values for t, but we are only interested in the positive value since time cannot be negative.
Using a calculator to simplify the equation, we find:
t ≈ 8.28 seconds (rounded to 2 decimal places)
Therefore, it will take approximately 8.28 seconds for the car to accelerate uniformly to a stop in exactly 102 m.