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Th equations to be used here are the following:
a = (v - v₀)/t
x = v₀t + 0.5at²
The speed of the fugitive is the sum of his own speed plus the speed of the train. Thus,
v₀ = 0 + 5.5 m/s = 5.5 m/s
v = 8 m/s + 5.5 m/s = 13.5 m/s
a.) We use the first equation to determine time
2.5 m/s² = (13.5 m/s - 5.5 m/s)t
Solving for t,
t = 3.2 seconds
b.) We use the answer in a) and the 2nd equation:
x = (5.5 m/s)(3.2 s) + 0.5(2.5 m/s²)(3.2 s)²
x = 30.4 meters
Answer:
The two objects will collide with the same position vector for all three components at exactly t = 4 s
Explanation:
For two particles starting out at the same time to collide, their position Vector's at the time of collision must be exactly the same.
So, at the collision point, position vector of object 1 is equated to that of object 2.
r₁ = (t², 13t-36, t²)
r₂ = (7t-12, t², 5t-4)
At he point of collision
t² = 7t - 12
t² - 7t + 12 = 0
t² - 4t - 3t + 12 = 0
t(t - 4) - 3(t - 4) = 0
t = 3s or t = 4s
13t - 36 = t²
t² - 13t + 36 = 0
t² - 4t - 9t + 36 = 0
t(t - 4) - 9(t - 4) = 0
t = 9s or 4s
t² = 5t - 4
t² - 5t + 4 = 0
t² - 4t - t + 4 = 0
t(t - 4) - 1(t - 4) = 0
t = 1s or t = 4s
The three components intersect at other times, but at t = 4s, they all intersect at the same time! Meaning that, at this point the two objects are at the same place with the same position vector at that time.
