Chapter 3 Problem 24

Problem

A particle starts from the origin at $t=0$ with an initial velocity of $5.0$ m/s along the positive $x$ axis. If the acceleration is $-3.0{\hat {i}}+4.5{\hat {j}}$ m/s$^{2}$, determine the velocity and position of the particle at the moment it reaches its maximum $x$ coordinate.

Solution

The acceleration is constant.

The particle reaches its maximum $x$ coordinate when the $x$ velocity is 0.

${\vec {r}}_{0}=0$

${\vec {v}}_{0}=5.0{\textrm {m/s}}{\hat {i}}$

${\vec {v}}={\vec {v}}_{0}+{\vec {a}}t=5.0{\hat {i}}{\textrm {m/s}}+(-3.0t{\hat {i}}+4.5t{\hat {j}}){\textrm {m/s}}$

$v_{x}=(5.0-3.0t){\textrm {m/s}}\rightarrow v_{xm}=0=(5.0-3.0t_{xm}){\textrm {m/s}}\rightarrow t_{xm}={\frac {5.0{\textrm {m/s}}}{3.0{\textrm {m/s}}^{2}}}=1.67{\textrm {s}}$

${\vec {v}}(t_{xm})=5.0{\hat {i}}{\textrm {m/s}}+\left(-3.0(1.67){\hat {i}}+4.5(1.67){\hat {j}}\right){\textrm {m/s}}=7.5{\textrm {m/s}}{\hat {j}}$

${\vec {r}}={\vec {r}}_{0}+{\vec {v}}_{0}t+{\frac {1}{2}}{\vec {a}}t^{2}=5.0t{\hat {i}}{\textrm {m}}+{\frac {1}{2}}\left(-3.0t^{2}{\hat {i}}+4.5t^{2}{\hat {j}}\right){\textrm {m}}$

${\vec {r}}(t_{xm})=5.0(1.67){\hat {i}}{\textrm {m}}+{\frac {1}{2}}\left[-3.0(1.67)^{2}{\hat {i}}+4.5(1.67)^{2}{\hat {j}}\right]{\textrm {m}}=4.2{\hat {i}}{\textrm {m}}+6.3{\hat {j}}{\textrm {m}}$