Animated Solution for Physics - Kinematics: Comprehension Passage
Two particle A and B are moving towards each other on a straight line with equal speeds 5 m/s. At an instant that is assumed t=0 s, distance between the particles is 100 m. It is desired to move another particle C always maintaining a distance 40 m from the particle A and 30 m from the particle B.
Question 1:
When and for how long can the particle C fulfil the given condition?
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* Multiple Correct
Question 2:
What is speed of the particle C at the instant t=5 s?
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* Multiple Correct
Question 3:
What is modulus of acceleration of the particle C at the instant t=5 s?
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* Multiple Correct
Question 4:
At the instant, when the line joining locations of A and B is perpendicular to the line joining locations of B and C, what are the magnitudes of velocities of C relative to A and B respectively?
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* Multiple Correct
Visualized Solution
KinematicSetup
xA(t)=5t
xB(t)=100−5t
AB=∣xB−xA∣=∣100−10t∣
TriangleInequality
∣AC−BC∣≤AB≤AC+BC
∣40−30∣≤∣100−10t∣≤40+30
10≤∣100−10t∣≤70
ValidTimeIntervals
Case 1: 100−10t>0⟹t<10
10≤100−10t≤70⟹3≤t≤9
Case 2: 100−10t<0⟹t>10
10≤10t−100≤70⟹11≤t≤17
t∈[3,9]∪[11,17]
CoordinatesofParticleC
AC2=(x−5t)2+y2=1600
BC2=(x−100+5t)2+y2=900
Subtracting gives: x=50+10−t35
Let u=10−t⟹x=50+u35
VelocityComponentsatt=5 s
vx=dtdx=u235=1.4 m/s
y2=1250−25u2−u21225⟹y=±24 m
2yvy=50u−u32450⟹vy=±4.8 m/s
vC=vx2+vy2=5 m/s
AccelerationComponentsatt=5 s
ax=dtdvx=u370=0.56 m/s2
vy2+yay=−25−u43675
(4.8)2+24ay=−25−5.88⟹ay=−2.2467 m/s2
aC=ax2+ay2≈2.3 m/s2
RelativeVelocityCondition
Line BC ⊥ Line AB⟹xC=xB
50+u35=50+5u⟹u2=7
vx=u235=5 m/s,vy=0 m/s
vCA=∣5−5∣=0 m/s,vCB=∣5−(−5)∣=10 m/s
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The Sigma Insight: Relative Velocity
Solution Diagram
The Kinematic Setup
Imagine two particles, A and B, rushing towards each other on a straight track.
We can anchor our entire analysis by setting up a simple coordinate system. Let's place particle A at the origin at t=0, moving to the right with a velocity of 5 m/s.
Particle B starts 100 m away and moves to the left at the exact same speed.
This means their positions at any time t are given by xA=5t and xB=100−5t.
The distance between them, which we'll call AB, is simply the absolute difference of their positions: AB=∣100−10t∣.
The Triangle Inequality Constraint
Now, enter particle C. It's playing a complex game of tag, forced to always stay exactly 40 m from A and 30 m from B.
For these three particles to coexist in a 2D plane, they must form a valid geometric triangle.
This brings us to the Triangle Inequality Theorem, which states that the length of any side of a triangle must be bounded by the difference and the sum of the other two sides.
Mathematically, this means ∣AC−BC∣≤AB≤AC+BC.
Substituting our known distances, we get a strict constraint on the distance between A and B: 10≤∣100−10t∣≤70.
Unlocking the Time Intervals
To find when particle C can actually exist, we need to solve this absolute value inequality.
We must split this into two distinct physical cases based on whether the particles have crossed paths yet.
Case 1: Before they cross (t<10 s), the distance is 100−10t.
Solving 10≤100−10t≤70 yields our first valid time window: 3≤t≤9 s.
Case 2: After they cross (t>10 s), the distance becomes 10t−100.
Solving 10≤10t−100≤70 gives us our second window: 11≤t≤17 s.
This perfectly answers our first question, confirming that C can fulfill the condition during the intervals 3 s≤t≤9 s and 11 s≤t≤17 s.
Decoupling the Coordinates
To find the speed and acceleration of C, we need its exact (x,y) coordinates.
We start by writing the distance squared equations for both AC and BC.
AC2=(x−5t)2+y2=1600
BC2=(x−100+5t)2+y2=900
Here is a beautiful algebraic trick: by subtracting the second equation from the first, the non-linear y2 terms completely vanish!
The x2 terms also cancel out, leaving us with a simple linear equation for x.
Solving this gives x=50+10−t35.
To make our upcoming calculus much cleaner, let's introduce a substitution: let u=10−t.
Our x-coordinate simplifies elegantly to x=50+u35.
Calculating Velocity at t=5 s
With our position equations ready, we can find the velocity components by differentiating with respect to time.
Remember that since u=10−t, its derivative is dtdu=−1.
Differentiating x gives the horizontal velocity: vx=u235.
At t=5 s, u is 5, which makes vx=1.4 m/s.
Next, we substitute x back into our first distance equation to find y2.
y2=1250−25u2−u21225
Instead of dealing with messy square roots, we use implicit differentiation to find the vertical velocity.
Differentiating both sides yields 2yvy=50u−u32450.
At t=5 s, we calculate y=±24 m, which gives us vy=±4.8 m/s.
The total speed is the magnitude of these components: vC=1.42+4.82.
This beautifully simplifies to exactly 5 m/s.
Calculating Acceleration at t=5 s
To find the acceleration, we must differentiate our velocity components one more time.
Differentiating vx gives the horizontal acceleration: ax=u370.
At t=5 s, this evaluates to 0.56 m/s2.
For the vertical acceleration, we differentiate our implicit velocity equation again.
vy2+yay=−25−u43675
Plugging in our known values of vy and y at t=5 s, we can solve for ay.
We find that ay≈−2.2467 m/s2.
The modulus of the total acceleration is aC=ax2+ay2.
Calculating this magnitude gives us approximately 2.3 m/s2.
The Perpendicular Relative Velocity
Finally, we analyze the specific instant when the line BC is perpendicular to AB.
Since AB lies perfectly horizontal along the x-axis, BC must be perfectly vertical.
This geometric constraint means that particles B and C must share the exact same x-coordinate: xC=xB.
Equating their position expressions gives 50+u35=50+5u.
Solving this yields u2=7.
At this exact turnaround moment, the vertical velocity vy becomes exactly zero.
The horizontal velocity vx evaluates to 735=5 m/s.
So, particle C is moving purely horizontally at 5 m/s.
Its velocity relative to A is ∣5−5∣=0 m/s, and relative to B is ∣5−(−5)∣=10 m/s.
