Question

In two dimensional motion, the position vector of a moving particle changes as : x(t) = 10 + 5t + 7t^2 and y(t) = 12 – 4t + 3t^2 Where x and y are in meter and t is in second. Calculate : (i) Distance from the origin at the initial state. (ii) Distance from origin 5 seconds later. (iii) Velocity and acceleration after 10 seconds.

Answer 1

Given:

In two dimensional motion, the position vector of a moving particle changes as : x(t) = 10 + 5t + 7t^2 and y(t) = 12 – 4t + 3t^2 Where x and y are in meter and t is in second.

To find:

(i) Distance from the origin at the initial state. (ii) Distance from origin 5 seconds later. (iii) Velocity and acceleration after 10 seconds.

Calculation:

At t = 0 ;

$x = 10 + 5t + 7 {t}^{2}$

$= > x = 10 + (5 \times 0) + 7 {(0)}^{2}$

$= > x = 10 \: m$

$y = 12 - 4t + 3 {t}^{2}$

$= > y = 12 -( 4 \times 0) + 3 {(0)}^{2}$

$= > y = 12 \: m$

So , initial distance from origin

$= \sqrt{ {(10)}^{2} + {(12)}^{2} }$

$= \sqrt{100 + 144}$

$= \sqrt{244}$

$\boxed{= 15.61 \: m}$

At t = 5 sec;

$x = 10 + 5t + 7 {t}^{2}$

$= > x = 10 + (5 \times 5) + 7 {(5)}^{2}$

$= > x = 10 + 25 + 175$

$= > x = 210 \: m$

$y = 12 - 4t + 3 {t}^{2}$

$= > y = 12 - (4 \times 5) + 3 {(5)}^{2}$

$= > y = 12 - 20+75$

$= > y = 67 \: m$

So, distance from origin

$= \sqrt{ {(210)}^{2} + {(67)}^{2} }$

$= \sqrt{44100 + 4489}$

$\boxed{ = 220.42 \: m}$

Velocity at t = 10 sec:

$v = \sqrt{ {(14t + 5)}^{2} + {(6t - 4)}^{2} }$

$= > v = \sqrt{ {(140 + 5)}^{2} + {(60 - 4)}^{2} }$

$= > v = \sqrt{ {(145)}^{2} + {(56)}^{2} }$

$= > v = \sqrt{ 24161}$

$\boxed{ = > v = 155.6 \: m {s}^{ - 1} }$

Acceleration at t = 10 secs:

$a = \sqrt{ {(14)}^{2} + {(6)}^{2} }$

$= > a = \sqrt{ 196 + 36}$

$= > a = \sqrt{232}$

$\boxed{ = > a = 15.23 \: m {s}^{ - 2} }$

Hope It Helps.