A police car stopped at a set of lights has a speeder pass it at 100 km/h. If the police car can accelerate at 3.6 m/s^2, how long does it take to catch the speeder?
They to the same distance in the same time.
d=100km/hr*t
d=1/2 3.6m/s^2 t
change km/hr to m/s in the first equation, then set d=d, and solve for t.
Why do you multiply by 1/2?
To find out how long it takes for the police car to catch the speeder, we need to determine the time it takes for the police car to reach the same speed as the speeder.
Let's start by converting the speed of the speeder from km/h to m/s. We know that 1 km/h is equal to 1000 m/3600 s. Therefore, the speeder's speed is 100 km/h * (1000 m/3600 s) = 27.78 m/s.
Next, we can calculate the time it takes for the police car to reach the same speed as the speeder using the acceleration of the police car.
We have an initial velocity (vi) of the police car, which can be assumed to be 0 m/s since it is stopped. The final velocity (vf) is the speed of the speeder, which is 27.78 m/s. The acceleration (a) is given as 3.6 m/s^2.
We can use the equation of motion s = vi * t + 0.5 * a * t^2, where s is the distance covered, vi is the initial velocity, a is the acceleration, and t is the time.
Since we want to find the time, we rearrange the equation to t = (vf - vi) / a.
Plugging in the values, we get t = (27.78 m/s - 0 m/s) / 3.6 m/s^2 ≈ 7.717 seconds.
Therefore, it takes approximately 7.717 seconds for the police car to catch the speeder.