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

(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)

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