Question

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)

Answer 1

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:-

$x(t)=5t-5$

$y(t)=2t-k$

Position of particles B:-

$x(t)=4t$

$y(t)=t^2-2t+1$

Part A)  For k = -6

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

Position of particle B, $(4t,t^2-2t-1)$

If both collides then x and y coordinate must be same

Therefore,

• For x-coordinate:

5t - 5 = 4t    

       t = 5

• For y-coordinate:

$2t+6=t^2-2t-1$

$t^2-4t-7=0$

$t=-1.3,5.3$

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:

$2t-k=t^2-2t-1$

Put t = 5

$10-k=25-10-1$

$k=4$

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

Part C)

Speed of particle A, $\dfrac{dA}{dt}$

$\dfrac{dA}{dt}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}$

$\dfrac{dA}{dt}=2\cdot \dfrac{1}{5}\approx 0.4$

Speed of particle B, $\dfrac{dB}{dt}$

$\dfrac{dB}{dt}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}$

$\dfrac{dB}{dt}=2t-2\cdot \dfrac{1}{4}$

At t = 5

$\dfrac{dB}{dt}=10-2\cdot \dfrac{1}{4}=2$

Hence, Particle B moves faster than particle A