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Two particles move in the xy-plane. At time t, the position of particle A is given by x(t)=5t−5 and y(t)=2t−k, and the position of particle B is given by x(t)=4t and y(t)=t2−2t−1.

(a) If k=−6, do the particles ever collide?

(b) Find k so that the two particles are certain to collide.

k=

(c) At the time the particle collide in (b), which is moving faster?
A. particle A
B. particle B
C. neither particle (they are moving at the same speed)

Respuesta :

Answer:

Part A)  Not collide

Part B)  k = 4

Part C)  Particle B is moving fast.

Step-by-step explanation:

Two particles move in the xy-plane. At time, t

Position of particle A:-

[tex]x(t)=5t-5[/tex]

[tex]y(t)=2t-k[/tex]

Position of particles B:-

[tex]x(t)=4t[/tex]

[tex]y(t)=t^2-2t+1[/tex]

Part A)  For k = -6

Position particle A, (5t-5,2t+6)

Position of particle B, [tex](4t,t^2-2t-1)[/tex]

If both collides then x and y coordinate must be same

Therefore,

  • For x-coordinate:

5t - 5 = 4t    

       t = 5

  • For y-coordinate:

[tex]2t+6=t^2-2t-1[/tex]

[tex]t^2-4t-7=0[/tex]

[tex]t=-1.3,5.3[/tex]

The value of t is not same. So, k = -6 A and B will not collide.

Part B) If both collides then x and y coordinate must be same

  • For x-coordinate:

5t - 5 = 4t    

       t = 5

  • For y-coordinate:

[tex]2t-k=t^2-2t-1[/tex]

Put t = 5

[tex]10-k=25-10-1[/tex]

[tex]k=4[/tex]

Hence, if k = 4 then A and B collide.

Part C)

Speed of particle A, [tex]\dfrac{dA}{dt}[/tex]

[tex]\dfrac{dA}{dt}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}[/tex]

[tex]\dfrac{dA}{dt}=2\cdot \dfrac{1}{5}\approx 0.4[/tex]

Speed of particle B, [tex]\dfrac{dB}{dt}[/tex]

[tex]\dfrac{dB}{dt}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}[/tex]

[tex]\dfrac{dB}{dt}=2t-2\cdot \dfrac{1}{4}[/tex]

At t = 5

[tex]\dfrac{dB}{dt}=10-2\cdot \dfrac{1}{4}=2[/tex]

Hence, Particle B moves faster than particle A

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