just see when the distances are equal:
120km/hr = 100/3 m/s, so
100/3 t = 2.5/2 t^2
t = 80/3
Now I expect you can handle the rest . . .
120km/hr = 100/3 m/s, so
100/3 t = 2.5/2 t^2
t = 80/3
Now I expect you can handle the rest . . .
Let's start by finding the time it takes for the police car to catch up to the speedster:
Step 1: Convert the speed of the speedster from km/h to m/s.
- Speed of the speedster = 120 km/h
- Speed of the speedster in m/s = 120 km/h * (1 km/3600 s) * (1m/1 km)
= (120 * 1000)/(3600 * 1) m/s
≈ 33.33 m/s
Step 2: Use the equation for the position of an object with constant acceleration:
- s = ut + (1/2)at^2
where s = distance, u = initial velocity, a = acceleration, and t = time.
Let's assume that the initial position of the police car is at point A, and the speedster is at point B when t = 0.
Step 3: Use the same equation for both the speedster and the police car. We can set up two separate equations:
- For the speedster:
s1 = 33.33t
- For the police car:
s2 = 0 + (1/2) * 2.5 * t^2
Step 4: Find the time at which the police car catches up to the speedster.
- When the police car catches the speedster, the distances covered by both are equal. Therefore, s1 = s2.
- 33.33t = (1/2) * 2.5 * t^2
- Simplifying, we get:
66.66t = 1.25t^2
- Rearranging, we get:
1.25t^2 - 66.66t = 0
- Factoring out t, we get:
t(1.25t - 66.66) = 0
- Solving for t, we find two solutions:
t = 0 (which is the starting point) and t ≈ 53.33s
Since time cannot be negative, we discard the solution t = 0.
So, the police car will take approximately 53.33 seconds to catch the speedster.
Step 5: Calculate the distance the speedster would have traveled when caught by the police car.
- Using the equation s1 = 33.33t, we plug in t = 53.33s:
s1 = 33.33 * 53.33
s1 ≈ 1777.78m
Therefore, the speedster would have traveled approximately 1777.78 meters when caught by the police car.