A car is moving down the road with a constant velocity of 20m/s. The car passes by a big billboard. There is a police officer ( motorcycle) at rest behind the billboard. It takes one second to start the motorcycle and start chasing the car. The motorcycle is accelerating at 2m/s^2. How long does it take for the police to catch up the car from the moment the motorcycle starts moving
if it takes t seconds after the cop gets going, then
20(t+1) = (1/2)(2)t^2
To find the time it takes for the police motorcycle to catch up with the car, we need to find the distance between them when the motorcycle starts moving and when they meet.
First, let's find the distance the car travels during the one-second delay after the motorcycle starts moving. Since the car is moving with a constant velocity of 20 m/s, the distance it covers in one second can be calculated using the formula:
Distance = Velocity × Time
Distance = 20 m/s × 1 s
Distance = 20 meters
So, during the one-second delay, the car travels 20 meters.
Now, the motorcycle starts from rest, so we can use the equation of motion to find the distance it travels when it catches up with the car. The equation of motion is:
Distance = Initial Velocity × Time + (1/2) × Acceleration × Time^2
Since the initial velocity of the motorcycle is 0 and the acceleration is given as 2 m/s^2, we can rewrite the equation as:
Distance = (1/2) × Acceleration × Time^2
We know that the distance the car travels during the one-second delay is 20 meters. Therefore, the distance the motorcycle has to travel to catch up with the car is also 20 meters.
Now we have the equation:
20 = (1/2) × 2 m/s^2 × Time^2
Simplifying the equation further:
20 = Time^2
Taking the square root of both sides:
√20 = Time
Calculating the square root of 20:
Time ≈ 4.47 seconds
Therefore, it will take approximately 4.47 seconds for the police motorcycle to catch up with the car from the moment the motorcycle starts moving.