Chapter 2 Problem 16

Problem

The position of a ball rolling in a straight line is given by $x=2.0-3.6t+1.1t^{2}$, where $x$ is in meters and $t$ in seconds.

(a) Determine the position of the ball at $t=$ 1.0 s, 2.0 s, and 3.0 s.

(b) What is the average velocity over the interval $t=$ 1.0 s to $t=$ 3.0 s?

(c) What is its instantaneous velocity at $t=$ 2.0s and at $t=$ 3.0s?

Solution

notice that if $x$ is in meters and and $t$ is in seconds, the implicit (hidden) units of the constants are $2.0\rightarrow 2.0{\textrm {m}}$; $3.6\rightarrow 3.6{\textrm {m/s}}$; $1.1\rightarrow 1.1{\textrm {m/s}}^{2}$

(a)

$x$(1.0s)= 2.0m−(3.6m/s)(1.0s)+(1.1m/s$^{2}$ )(1.0s)$^{2}$ =-0.5 m

$x$(2.0s)= 2.0m−(3.6m/s)(2.0s)+(1.1m/s$^{2}$ )(2.0s)$^{2}$ =-0.8 m

$x$(3.0s)= 2.0m−(3.6m/s)(3.0s)+(1.1m/s$^{2}$ )(3.0s)$^{2}$ = 1.1 m

(b)

${\bar {v}}={\frac {\Delta x}{\Delta t}}={\frac {1.1{\textrm {m}}-(-0.5{\textrm {m}})}{2.0{\textrm {s}}}}=0.80{\textrm {m/s}}$

(c)

$v(t)={\frac {dx(t)}{dt}}=-3.6{\textrm {m/s}}+1.1{\textrm {m/s}}^{2}t$

Failed to parse (syntax error): {\displaystyle v(2.0\textrm{s})=−3.6\textrm{m/s}+(2.2\textrm{m/s}^2 )(2.0\textrm{s})=0.8\textrm{m/s}}

Failed to parse (syntax error): {\displaystyle v(3.0\textrm{s})=−3.6\textrm{m/s}+(2.2\textrm{m/s}^2 )(3.0\textrm{s})=3.0\textrm{m/s}}