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/s2,
a) how long does it take to catch the speeder?
b) how far would the police car have to go before it catches the speeder?
c) what would its speed be when it caught up with the car? Is this speed reasonable?
I know the answers are a) 15 s, b) 427 m, c) 55 m/s but I don't know how to get these answers please help
To solve this problem, we can use the equations of motion and a few basic principles of physics.
a) To find the time it takes for the police car to catch the speeder, we can assume both vehicles start from rest. We'll use the equation: distance = initial velocity × time + (1/2) × acceleration × time^2.
In this case, the distance covered by the police car is the same as the distance covered by the speeder. Let's assign variables:
- d = distance covered by both vehicles
- v = initial velocity of the police car (0 km/h)
- v_s = speed of the speeder (100 km/h = 100,000 m/3600 s)
- a = acceleration of the police car (3.6 m/s^2)
- t = time taken to catch the speeder (unknown)
Using the equation, we have:
d = v × t + (1/2) × a × t^2
d = 0 × t + (1/2) × 3.6 × t^2
d = 1.8 t^2
Since both vehicles cover the same distance, the distance covered by the speeder is also d. We can write this as:
d = v_s × t + (1/2) × 0 × t^2
d = (100000/3600) × t + 0
d = 27.78 t
Now, equating the two expressions for d, we get:
1.8 t^2 = 27.78 t
Rearrange this equation by subtracting 27.78 t from both sides:
1.8 t^2 - 27.78 t = 0
Now we have a quadratic equation that we can solve to find t. Divide both sides by t to get:
1.8 t - 27.78 = 0
Simplifying, we get:
1.8 t = 27.78
Finally, solve for t:
t = 27.78 / 1.8 ≈ 15.43 seconds
b) To find the distance the police car would have to go before catching the speeder, we can substitute the value of t into either d = 1.8 t^2 or d = 27.78 t.
Using d = 1.8 t^2:
d = 1.8 × (15.43)^2
d ≈ 423.96 meters
So, the police car would have to go approximately 423.96 meters before catching the speeder.
c) To find the speed of the police car when it catches up with the speeder, we can use the equation: final velocity = initial velocity + (acceleration × time).
The initial velocity of the police car is zero, and the time is approximately 15.43 seconds. The acceleration is given as 3.6 m/s^2.
Using the equation, we have:
final velocity = 0 + (3.6 × 15.43)
final velocity ≈ 55.55 m/s
Therefore, the speed of the police car when it catches the speeder would be approximately 55.55 m/s.
In terms of reasonability, this speed can be considered reasonable for a police car chasing a speeder, as it is within the typical range of speeds for police vehicles.