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At time t=0 particle A is at x= -100m, moving with velocity 20m/s along positive x- axis particle B is at y= -84m, moving with velocity 25m/s along positive y -axis . Distance of closest approach is?
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At time t=0 particle A is at x= -100m, moving with velocity 20m/s alon...
Both the particles will always be right angled to each other irrespective of their velocities at all times. So, the separation will be the
hypotenuse of the triangle formed by the origin and the two particles.
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At time t=0 particle A is at x= -100m, moving with velocity 20m/s alon...
Problem Statement:
At time t=0 particle A is at x= -100m, moving with velocity 20m/s along positive x- axis particle B is at y= -84m, moving with velocity 25m/s along positive y -axis . Distance of closest approach is? Explain in details.
Solution:
To find the distance of closest approach we have to first find the time at which the particles are closest to each other. We can then plug in this time to find the distance of closest approach.
Step 1: Find the time of closest approach
Let's assume that the particles are closest to each other at time t. At this time, the x-coordinate of particle A and the y-coordinate of particle B will be equal. We can write the following equation to find t:
xA(t) = x0A + vAt
yB(t) = y0B + vBt
where x0A = -100m, vA = 20m/s, y0B = -84m, and vB = 25m/s.
Setting these two equations equal to each other and solving for t, we get:
-100m + 20m/s * t = 25m/s * t - 84m
t = 6.4s
Therefore, the particles are closest to each other at time t = 6.4s.
Step 2: Find the distance of closest approach
Now that we know the time of closest approach, we can find the distance of closest approach using the following equation:
d = sqrt((xA(t) - xB(t))2 + (yA(t) - yB(t))2)
where xB(t) and yA(t) are the positions of particles B and A at time t, respectively.
Plugging in the values, we get:
d = sqrt((-100m + 20m/s * 6.4s)2 + (-84m + 25m/s * 6.4s)2)
d = 104.8m
Therefore, the distance of closest approach is 104.8m.
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At time t=0 particle A is at x= -100m, moving with velocity 20m/s along positive x- axis particle B is at y= -84m, moving with velocity 25m/s along positive y -axis . Distance of closest approach is?
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