Animated Solution for Physics - Kinematics: A particle is moving along the x-axis with its coordinate with the t given by x(t)=10+8t−3t2. Another particle is moving along the y-axis with its coordinate as a function of time given by y(t)=5−8t3. At t=1 s, the speed of the second particle as measured in the frame of the first particle is given as v. Then v (in m/s) is ......... .
Enter Numerical Value:
Visualized Solution
Position Equations
x(t)=10+8t−3t2
y(t)=5−8t3
Velocity from Position
v=dtdr
vA=dtdxi^
vB=dtdyj^
Velocity of Particle A
vA=dtd(10+8t−3t2)i^
vA=(8−6t)i^
Velocity of A at t=1 s
At t=1 s:
vA=(8−6(1))i^
vA=2i^ m/s
Velocity of Particle B
vB=dtd(5−8t3)j^
vB=(−24t2)j^
Velocity of B at t=1 s
At t=1 s:
vB=−24(1)2j^
vB=−24j^ m/s
Relative Velocity Concept
Velocity of B with respect to A:
vBA=vB−vA
Calculating Relative Velocity
vBA=−24j^−(2i^)
vBA=−2i^−24j^ m/s
Magnitude of Relative Velocity
∣vBA∣=(−2)2+(−24)2
∣vBA∣=4+576
∣vBA∣=580 m/s
Final Answer
Given: ∣vBA∣=v
580=v
v=580
The Way Forward
What if we needed the relative acceleration?
aBA=aB−aA
00:00 / 00:00
The Sigma Insight: Relative Velocity
Solution Diagram
The Dance of Two Particles
Imagine you are standing at the origin of a massive coordinate system. You see two particles, let's call them Particle A and Particle B, moving along the axes. Particle A is constrained to the X-axis, performing a dance dictated by the equation x(t)=10+8t−3t2. Meanwhile, Particle B is sliding along the Y-axis, following its own rhythm given by y(t)=5−8t3.
The question asks us for a very specific perspective: what is the speed of Particle B if you were sitting on Particle A at the exact moment t=1 s? This is a classic relative motion problem, and to solve it, we first need to figure out how fast each particle is moving individually.
Unlocking the Velocities
In kinematics, position is just the starting point. To find the velocity, we need to look at the rate of change of position. Mathematically, this means taking the derivative of the position equations with respect to time.
Let's start with Particle A. Its position is x(t)=10+8t−3t2. Differentiating this gives us its velocity:
vA=dtdx=8−6t
Since Particle A is moving along the X-axis, its velocity vector is vA=(8−6t)i^.
Now, let's look at Particle B. Its position is y(t)=5−8t3. Differentiating this gives us:
vB=dtdy=−24t2
Since Particle B is moving along the Y-axis, its velocity vector is vB=−24t2j^.
Freezing Time at t=1 s
The problem asks for the relative speed at a specific instant: t=1 s. Let's plug this time into our velocity equations to see exactly what's happening at that moment.
For Particle A:
vA=8−6(1)=2 m/s
So, vA=2i^ m/s. Particle A is moving to the right at a leisurely pace.
For Particle B:
vB=−24(1)2=−24 m/s
So, vB=−24j^ m/s. Particle B is zooming downwards along the Y-axis!
The Relative Perspective
Now comes the crucial step. We need the velocity of Particle B as measured in the frame of Particle A. This is the relative velocity, denoted as vBA. The formula for relative velocity is beautifully simple:
vBA=vB−vA
Substituting our vectors into this equation:
vBA=−24j^−(2i^)=−2i^−24j^ m/s
This vector tells us that if you were sitting on Particle A, you would see Particle B moving to the left at 2 m/s and downwards at 24 m/s.
Calculating the Final Speed
The question doesn't just ask for the velocity vector; it asks for the speed, which is the magnitude of the velocity vector. To find the magnitude of vBA, we use the Pythagorean theorem:
∣vBA∣=(−2)2+(−24)2
∣vBA∣=4+576
∣vBA∣=580 m/s
The problem states that this speed is equal to v. By comparing our result with the given expression, we can easily see that:
580=v
Therefore, the value of v is exactly 580.
This problem beautifully illustrates how calculus and vector algebra come together to solve relative motion scenarios. Always remember to find the individual velocity vectors first before jumping into the relative frame!