In the same instant that a Porsche races past him at 130 km/h, a speed cop gets off his motorcycle, accelerates uniformly at 8 m/s^2, to catch the speedster. After 5 seconds the trafic officer reaches his top speed, which he then maintains until he reaches up to the car. How far down the road does the bike draw level with the Porsche?
The cop should really get ON the motorcycle if he wants to catch the guy. Which he never will because his top speed is only 40 kph.
Oops. My mistake, the acc is in m/s^2.
So the porsche is going 36.1 m/s.
x = 180.5 + 36.1t (180.5 during 5 seconds)
and for the cop
x = 100 + 40t (the 100 is how far he went during the acceleration)
so 3.9t = 80.5
and add t = 20.6 + 5 =25.6
x = 25.6*36.1
To determine how far down the road the bike draws level with the Porsche, we need to break down the problem into smaller steps. Let's calculate the necessary information step by step:
1. Convert the speed of the Porsche from km/h to m/s.
- 130 km/h = (130 * 1000) / 3600 m/s = 36.11 m/s
2. Find the time it takes for the motorcycle officer to reach his top speed.
- The officer accelerates uniformly at 8 m/s^2 until he reaches his top speed.
- We can use the equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
- Here, the initial velocity (u) is 0 m/s (since the officer is starting from rest).
- The acceleration (a) is 8 m/s^2.
- The final velocity (v) is unknown, but it is the top speed of the officer.
- We can rearrange the equation to solve for time (t): t = (v - u) / a.
- Plugging in the values, we get: t = (v - 0) / 8.
- Since it takes 5 seconds for the officer to reach his top speed, we can equate the two time values: (v - 0) / 8 = 5.
- Solving for v, we get: v = 8 * 5 = 40 m/s.
3. Calculate the distance covered by the motorcycle during the acceleration phase.
- To find this distance, we can use the equation: s = ut + (1/2)at^2, where s is the distance covered, u is the initial velocity, a is the acceleration, and t is the time taken.
- Here, the initial velocity (u) is 0 m/s (since the officer starts from rest).
- The acceleration (a) is 8 m/s^2.
- The time taken (t) is 5 seconds (as given in the question).
- Plugging in the values, we get: s = 0 * 5 + (1/2) * 8 * (5^2).
- Solving for s, we get: s = 0 + (1/2) * 8 * 25 = 100 m.
4. Determine the distance covered by both the Porsche and the motorcycle once the officer reaches his top speed.
- Since the officer maintains his top speed until he catches up to the Porsche, the distance covered by both vehicles will be the same.
- We need to find the time taken by both vehicles to cover the equal distance, using the equation: distance = speed * time.
- Let's assume the time taken by both vehicles to cover the distance is t. The distance covered by the motorcycle (d) is given by d = (40 m/s) * t.
- Similarly, the distance covered by the Porsche will be equal to the distance it travels in 5 seconds plus the distance it covers at its top speed.
- The distance covered by the Porsche (d) is given by d = (36.11 m/s * 5 s) + (36.11 m/s * t).
5. Equate the distances covered by the Porsche and the motorcycle.
- We can set the two equations equal to each other and solve for t (time):
40t = (36.11 * 5) + (36.11 * t).
- Solving for t, we get:
40t - 36.11t = (36.11 * 5).
3.89t = 180.55.
t = 180.55 / 3.89 = 46.45 seconds.
6. Calculate the final distance covered by the motorcycle.
- Now that we have found the time (t), we can calculate the distance (d) covered by the motorcycle during this time.
- d = (40 m/s) * t = 40 * 46.45 = 1858 m.
Therefore, the motorcycle will draw level with the Porsche after traveling approximately 1858 meters down the road.