Question

The equations of motion of a rocket are \[x = 2t,\; y = - 4t,\; z = 4t,\] where the time t is given in seconds and the coordinate of a moving point in kilometres. At what distance will the rocket be from the starting point O (0, 0, 0) in 10 seconds?

Answer 1

Hint: We start with our equation of motion which is given in terms of t itself. And we are also starting our journey from the origin so using the given time we can easily calculate the distance required.

Complete step by step solution: We are given the equation of motion of the rocket, \[x = 2t,y = - 4t,z = 4t,\]
Thus the path of the Rocket represents a straight line passing through the origin (0, 0, 0 ) [for t=0, the equation of motion goes through the centre] for t=10sec.

So, we have, \[x = 2 \times 10,y = - 4 \times 10,z = 4 \times 10\] or, \[x = 20,y = - 40,z = 40\]

Therefore the position vector\[(\vec r)\] of the rocket will be given by
\[\therefore \vec r = x\hat i + y\hat j + z\hat k\]
And the distance between a point (a,b,c) and vector \[(\vec r)\] is given by \[\left| r \right| = \sqrt {{{(x - a)}^2} + {{(y - b)}^2} + {{(z - c)}^2}} \]
 So the distance of the rocket at t=10sec from the starting point i.e. (0,0,0)
⟹\[\left| r \right| = \sqrt {{{(x - 0)}^2} + {{(y - 0)}^2} + {{(z - 0)}^2}} \]
\[ \Rightarrow \left| r \right| = \sqrt {{{(20 - 0)}^2} + {{( - 40 - 0)}^2} + {{(40 - 0)}^2}} \]

\[ \Rightarrow \left| r \right| = \sqrt {400 + 1600 + 1600} \]
\[ \Rightarrow \left| r \right| = \sqrt {3600} \]\[ = 60km\]
since\[\vec r = x\hat i + y\hat j + z\hat k\]

\[\therefore \vec r = 2t\hat i - 4t\hat j + 4t\hat k\]

Note: Here we had used the formula of the distance of two points to find the solution. And if practically speaking, the rocket has the equations of motion given from the starting and at t=0 the starting point is the origin. So, we are starting it from the origin to proceed.