Question

When t = 0, a particle at (1,0,0) moves towards (4,4,12) with a constant speed of 65 ms. The position of the particle is measured in meter and the time in second. Assume constant velocity, the position of the particle for t=2 s is​

Answer 1

velocity change in accelration

Answer 2

Answer:

distance between points(1,0,0) and (4,4,12)=√(4-1)²+(4-0)²+(12-0)²=√3²+4²+12²

=√9+16+144=√169=13 units

velocity=65m/sec

in 2 sec ,distance covered=65×2=130m

let (x,y,z) be the position of particle after t=2 sec

then,√(x-1)²+(y-0)²+(z-0)²=130

√x²+1-2x+y²+z²=130

x²+y²+z²-2x=130²-1

x²+y²+z²-2x=16900-1=16899

x²+y²+z²=16899+2x

√(x-4)²+(y-4)²+(z-12)²=130-13

x²+16-8x+y²+16-8y+z²+144-24z=127²

x²+y²+z²-8x-8y-24z+176=16129

-8(x+y+3z)+16899+2x+176=16129

-8(x+y+3z)+2x=16129-17075

-6x-8y-24z=-946

-2(3x+4y+12z)=-946

3x+4y+12z=-946/-2=473...(1)

equation of line passing through (x1,y1,z1) and(x2,y2,z2) is (x-x1)/(x2-x1)=(y-y1)/y2-y1)=(z-z1)/z2-z1

(x-1)/(4-1)=(y-0)/(4-0)=(z-0)/(12-0)

(x-1)/3=y/4=z/12

now,y/4=(x-1)/3

y=4(x-1)/3=(4x-4)/3

z/12=(x-1)/3

z=4(x-1)=4x-4

put value of y,z in (1)

3x+4(4x-4)/3+12(4x-4)=473

3x+16x-16+48x-48=473

67x=473+64

x=537/67

y=(4(537/67)-4)/3

=2148-268/201=1880/201

z=4(537/66)-4=2148-268/67=1880/67

so (x,y,z)=(537/67,1880/201,1880/67) is the position of particle after t=2 sec