Answer to Question #279608 in Mechanics | Relativity for Malefti

Question

A particle moves along a straight line with constant acceleration a=3 m/s2

. At t=0 the

position of the particle is x0=3 m and the particle is at rest (v0=0).

i) Find the position and the velocity of the particle at t=2 s [3 mks]

ii) Find the position of the particle when v=30 m/s [3 mks]

Answer 1

The information given is

The initial position is $X_{\circ}=0\;m$

The initial velocity is $V_{\circ}=0 \; m/s$

The acceleration is $a=3\;m/s^{2}$

Part i

The position is given by

$X=X_{\circ}+V_{\circ}\;t+\dfrac{1}{2}\;a\;t^{2}$

Evaluating numerically in t=2 s

$X=X_{\circ}+V_{\circ}\;t+\dfrac{1}{2}\;a\;t^{2}\\ X=0\;m+0\;m/s\times 2\;s+\dfrac{1}{2}\times 3\;m/s^{2}\times (2\;s)^{2}\\ X=6\;m$

The position at t-2 s is $X=6\;m$

The velocity is given by

$V=V_{\circ}+a\;t$

Evaluating numerically.

$V=V_{\circ}+a\;t\\ V=0\;m/s+3\;m/s\times 2\;s\\ V=6\;m/s$

The velocity at t= 2s is $V=6\;m/s$

Part ii

The velocity is given by

$V=V_{\circ}+a\;t$

Obtaining the expression for the time.

$V=V_{\circ}+a\;t\\\\ V_{\circ}+a\;t=V\\\\ a\;t=V-V_{\circ}\\\\ t=\dfrac{V-V_{\circ}}{ a}\\\\$

Evaluating numerically.

$t=\dfrac{30\;m/s-0\;m/s}{ 3\;m/s^{2}}\\\\ t=10\;s$

The time where the velocity is 30 m/s is $t=10\;s$

The position is given by 

$X=X_{\circ}+V_{\circ}\;t+\dfrac{1}{2}\;a\;t^{2}$

Evaluating numerically.

$X=0\;m+0\;m/s\times 10\;s+\dfrac{1}{2}\times 3\;m/s^{2}\times (10\;s)^{2}\\ X=150\;m$

The position when V=30 m/s is $X=150\;m$