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Question: Two particles A and B are moving in the XY-plane. Their positions vary with time t according to rela...

Two particles A and B are moving in the XY-plane. Their positions vary with time t according to relation
XA(t)=3t{X_A}(t) = 3t , XB(t)=6{X_B}(t) = 6
YA(t)=t{{\text{Y}}_A}(t) = t , YA(t)=2+3t2{{\text{Y}}_A}(t) = 2 + 3{t^2}
Distance between two particles at t=1t = 1is:
A) 55
B) 33
C) 44
D) 12\sqrt {12}

Explanation

Solution

In order to solve the question, we will first find the coordinates at which the two particles are located at t=1t = 1. Then, we find the distance between the two particles using the distance formula as (XA−XB)2+(YA−YB)2\sqrt {{{\left( {{X_A} - {X_B}} \right)}^2} + {{\left( {{Y_A} - {Y_B}} \right)}^2}} , where XA,XB,YA,YB{X_A},{X_B},{Y_A},{Y_B} are the coordinates in XY plane.

Complete answer:
In the question we are given four different equations which show the relation of x coordinate and y coordinate with time t and we are also given the value of time at which we have to find the distance
Time (t) = 1 sec
Now we will substitute the value of t in the different equation
Equation 1
XA(t)=3t{X_A}(t) = 3t
Substituting t=1t = 1
XA=3{X_A} = 3
Equation 2
XB(t)=6{X_B}(t) = 6
Equation 3
YA(t)=t{{\text{Y}}_A}(t) = t
Substituting t=1t = 1
YA=1{{\text{Y}}_A} = 1
Equation 4
YA(t)=2+3t2{{\text{Y}}_A}(t) = 2 + 3{t^2}
Substituting t=1t = 1
YA=5{{\text{Y}}_A} = 5
Now, we have calculated all the coordinates independent of time.
Now, we can substitute the value of XA{X_A}, XB{X_B}, YA{Y_A} and YB{Y_B} into the distance formula to find the distance between the two particles at t=1t = 1.
So, we have,
Distance AB = (XA−XB)2+(YA−YB)2\sqrt {{{\left( {{X_A} - {X_B}} \right)}^2} + {{\left( {{Y_A} - {Y_B}} \right)}^2}}
Substituting the known values, we get,
Distance AB =(3−6)2+(1−5)2= \sqrt {{{\left( {3 - 6} \right)}^2} + {{\left( {1 - 5} \right)}^2}}
After solving squares and opening bracket, we get,
Distance AB =(−3)2+(−4)2= \sqrt {{{\left( { - 3} \right)}^2} + {{\left( { - 4} \right)}^2}}
=9+16= \sqrt {9 + 16}
=25= \sqrt {25}
Now, we know that 2525 is a square of five. So, we get,
=5= 5
So, the distance between the two objects at t=1t = 1 is five units.

Hence, option (A) is the correct answer.

Note:
The distance formula used in the question is applicable only for two dimensional geometry. Since it was mentioned that the two given particles move in the xy plane only, hence we have used this distance formula. In three dimensional geometry, we use an extended form of distance formula taking into account the z coordinates of the point as: (XA−XB)2+(YA−YB)2+(ZA−ZB)2\sqrt {{{\left( {{X_A} - {X_B}} \right)}^2} + {{\left( {{Y_A} - {Y_B}} \right)}^2} + {{\left( {{Z_A} - {Z_B}} \right)}^2}} .