PHYS 101: Solutions for Test 1

PHYS 101 - Physics

Test 1 Solutions

The converted measurement units are:
1. One meter constitutes $100 cm$ :
  $\begin{matrix} 9,385 {cm}^{3} & = & (9,385 {cm}^{3}) {(\frac{1 m}{100 cm})}^{3} \\ = & (9,385 {cm}^{3}) (\frac{1 m^{3}}{10^{6} {cm}^{3}}) \\ = & 9,385 \times 10^{- 6} m^{3} & = & 9.385 \times 10^{- 3} m^{3} & . \end{matrix}$
2. $1.468 \times 10^{- 2} kg$ .
3. Because there are $2.47$ acres in one hectare:
  $\begin{matrix} 289.6 ha & = & (289.6 ha) (\frac{2.47 acre}{1 ha}) & \approx & 7.15 \times 10^{2} acre & . \end{matrix}$
4. Using $0.3048$ meter per one foot:
  $\begin{matrix} 6,087.0 ft & = & (6,087.0 ft) (\frac{0.3048 m}{1 ft}) & \approx & 1.855 \times 10^{3} m & . \end{matrix}$
The correct defining phrases for the four formulas are:
1. average acceleration,
2. instantaneous acceleration,
3. instantaneous velocity,
4. average velocity.
An object is in free fall whenever its state of motion is influenced exclusively by gravity.

The accuracy estimates for the numbers are:

	number	s.f.	dec.
a.	$7 \times 10^{21}$	$1$	$- 21$
b.	$- 0.458 \times 10^{- 8}$	$3$	$11$
c.	$0.0018$	$2$	$4$
d.	$9.36$	$3$	$2$

The two vectors may point in the same direction, in which case the result is a vector of length seventeen. They might also have opposite directions, so that the result is a vector of length one. The correct answer is alternative c.
The resultant force on the curling stone is:
$\begin{matrix} F & = & [\begin{matrix} 90.1 \frac{(kg \cdot m)}{s^{2}} \\ - 162.2 \frac{(kg \cdot m)}{s^{2}} \end{matrix}] & . \end{matrix}$
If you want to describe the force in terms of magnitude and angle; this is a force of strength $|F| = 185.5 \frac{(kg \cdot m)}{s^{2}}$ , directed at an angle $θ_{v} = - 60.9 °$ from the positive $x$ -axis.
The two graphs appearing on this question were adapted from the ones appearing in the analysis of the simplified squirrel motion on slide 15 of Chapter 2.
1. In the first graph, the object moves forward for the second time beginning at approximately $t = 0.94 s$ .
2. Acceleration of the object can be inferred from the second graph. The slope of that graph is negative at $t = 0.70 s$ , and so acceleration in that neighborhood is negative.
This problem asks for the $x$ - and $y$ -components of the vector $N$ .

The component $N_{x} = 4.9 \cos 105 ° \approx - 1.3 \frac{m}{s^{2}}$ , and $N_{y} = 4.9 \sin 105 ° \approx 4.7 \frac{m}{s^{2}}$ .
This problem asks us to compute the magnitude and direction for the vector $V$ . The magnitude is $|V| = {[{(- 6.2)}^{2} + {(- 4.6)}^{2}]}^{\frac{1}{2}} = \sqrt{59.6} \approx 7.7$ . The direction of the vector is an angle measured counter-clockwise from the positive $x$ -axis: $θ_{V} = Arctan [\frac{(- 4.6)}{(- 6.2)}] + 180 ° \approx 1.4 \times 10^{2}$ degrees.
Because the launch site and the landing area are at substantially different heights, the formula for projectile range cannot be used here.
1. The inital velocity vector for the stone is ${[\begin{matrix} v_{x_{0}} & v_{y_{0}} \end{matrix}]}^{T}$ $=$ ${[\begin{matrix} v_{0} \cos θ & v_{0} \sin θ \end{matrix}]}^{T}$ $\approx$ ${[\begin{matrix} 3.5356 & 3.5356 \end{matrix}]}^{T}$ . The equation for horizontal motion of the stone is $x (t) \approx 3.5356 t + 0.0000$ , and the equation for vertical motion is $y (t) \approx - 4.90 t^{2} + 3.5356 t + 11.5$ .
  
  Setting $y (t) = 0.000$ in the vertical motion equation, a solution for $t$ indicates that the stone hits the ground at $t_{G} \approx 1.93 s$ .
2. The equations for the changing velocity of the stone are $v_{x} (t) \approx 3.5356$ and $v_{y} (t) \approx - 9.80 t + 3.5356$ . Evaluating these at $t_{G} \approx 1.93 s$ gives ${[\begin{matrix} v_{x} & v_{y} \end{matrix}]}^{T}$ $\approx$ ${[\begin{matrix} 3.54 \frac{m}{s} & - 15.4 \frac{m}{s} \end{matrix}]}^{T}$ .
Represented as a column vector, the initial velocity of the particle is:
$\begin{matrix} v_{0} & = & [\begin{matrix} 0.0 \\ 4.0 \end{matrix}] & . \end{matrix}$
When we track the velocity as it varies with time, the velocity equation
$\begin{matrix} v (t) & = & [\begin{matrix} 1.5 \\ 2.6 \end{matrix}] t & + & [\begin{matrix} 0.0 \\ 4.0 \end{matrix}] \end{matrix}$
computes any velocity vector that we might need. Specifically,
$\begin{matrix} v (2.6) & \approx & [\begin{matrix} 3.9 \frac{m}{s} \\ 10.8 \frac{m}{s} \end{matrix}] & . \end{matrix}$
The time that has elapsed during the acceleration in not directly known. Only the distance that was covered is clearly stated: $125 m$ .

Using $v_{0} = 3.5 \frac{m}{s}$ , $v = 17.2 \frac{m}{s}$ , and (arbitrarily) setting $x_{0} = 0.0 m$ at the beginning of the time in which the acceleration occurs, our one formula that directly links position and acceleration (with no time variable $t$ ) is:
$\begin{matrix} v^{2} & = & v_{0}^{2} & + & 2 a (x - x_{0}) & , \\ {(17.2)}^{2} & = & {(3.5)}^{2} & + & 2 a (125) & . \end{matrix}$
Now we just need to solve the equation depicted above, to get the value $a \approx 1.13 \frac{m}{s^{2}}$ .
The boat exhibits a different behavior for its seven hundred fifty meter trip upstream, than it does for the seven hundred fifty meter trip down. When traveling upstream, the velocity is $v_{up} = 5.5 - 2.5 = 3.0 \frac{m}{s}$ . Traveling downstream, the velocity of the boat becomes $v_{down} = 5.5 + 2.5 = 8.0 \frac{m}{s}$ .

To get travel times for the two seven hundred fifty meter segments, we just use the formula $x = v t$ twice. In each situation, $x = 750 m$ . Using the two different velocities $v_{up}$ and $v_{down}$ in the distance/rate/time equation, we get times $t_{up} = 250 s$ and $t_{down} = 93.75 s$ . Altogether, the trip is $t_{total} \approx 340 s$ .