Answer :
Answer:
First, let's find the time it takes for truck C to reach the location of truck A, since truck B wants to cross truck A before truck C arrives.
The relative velocity of truck C with respect to truck A is:
v(CA) = v(C) - v(A) = 15 m/s - 10 m/s = 5 m/s
The time it takes for truck C to reach the location of truck A is:
t(CA) = d(CA) / v(CA) = 500 m / 5 m/s = 100 s
Now, let's find the time it takes for truck B to cross truck A. Since they are moving in opposite directions, their relative velocity is:
v(BA) = v(B) + v(A) = 15 m/s + 10 m/s = 25 m/s
The time it takes for truck B to cross truck A is:
t(BA) = d(BA) / v(BA) = 500 m / 25 m/s = 20 s
Since truck B wants to cross truck A before truck C arrives, it must have crossed truck A within 100 seconds. Therefore, let's find the minimum acceleration required for truck B to cross truck A in less than 100 seconds.
The distance truck B travels while crossing truck A is:
d(B) = v(B) * t(BA) + 0.5 * a * t(BA)^2
We can solve for the minimum acceleration required by using the fact that truck B must cross truck A before truck C arrives, i.e., t(BA) + t(CA) <= 100 s.
Substituting the values we have found so far, we get:
500 m = 15 m/s * 20 s + 0.5 * a * (20 s)^2
240 m = 0.5 * a * (20 s)^2
a = 240 m / (20 s)^2 = 0.6 m/s^2
However, this acceleration is not sufficient for truck B to cross truck A in the 7th second and have a displacement of 24 meters. Let's find the acceleration required for truck B to reach a displacement of 24 meters by the 9th second.
Since truck B travels at a constant velocity of 15 m/s for the first 20 seconds, it covers a distance of:
d(1) = v(B) * t(1) = 15 m/s * 20 s = 300 m
The remaining distance for truck B to cover in the next 2 seconds (7th to 9th second) is:
d(2) = 500 m - 300 m = 200 m
In the 2 seconds, truck B must cover the remaining distance with an acceleration of:
a = (v(f)^2 - v(i)^2) / (2 * d(2))
where v(i) is the initial velocity of truck B, v(f) is its final velocity, and d(2) is the remaining distance.
We have:
v(i) = 15 m/s
d(2) = 200 m
and we want to find v(f) and a.
We can write the equation of motion:
d(2) = 0.5 * a * t(2)^2 + v(i) * t(2)
where t(2) is the time it takes for truck B to cover the remaining distance.
Since d(2) = 200 m, t(2) = 2 s, and v(i) = 15 m/s, we can solve for v(f):
200 m = 0.5 * a * (2 s)^2 + 15 m/s * 2 s
v(f) = sqrt(200 m - 4 a) + 15 m/s
Substituting the expression for v(f) into the equation for a, we get:
a = (v(f)^2 - v(i)^2) / (2 * d(2))
a = [(sqrt(200 m - 4 a) + 15 m/s)^2 - (15 m/s)^2] / (2 * 200 m)
Solving for a, we get:
a = 0.687 m/s^2
Therefore, the minimum acceleration required for truck B to cross truck A before truck C arrives is 0.6 m/s^2, and the acceleration required for truck B to reach a displacement of 24 meters by the 9th second is approximately 0.687 m/s^2.