A body can be treated as a point object when its size is negligible compared with the distance or scale of motion, and its rotation or tumbling is not important. The carriage and monkey-cyclist system can be approximated as point objects for large-scale translational motion. The spinning ball and tumbling beaker cannot, because rotational motion and size are relevant.
a. Yes; b. Yes, if the track size is large compared with the monkey-man system; c. No; d. No.
Time to office $=2.5/5=0.5\,\text{h}=30\,\text{min}$. Return time $=2.5/25=0.1\,\text{h}=6\,\text{min}$. Thus the $x$-$t$ graph has a positive straight segment, a horizontal rest segment, and a steeper negative straight segment.
The graph rises linearly from $(9.00\,\text{am},0)$ to $(9.30\,\text{am},2.5\,\text{km})$, stays horizontal until $5.00\,\text{pm}$, and then falls linearly to $(5.06\,\text{pm},0)$.
In each cycle of $5$ forward and $3$ backward steps, time $=8\,\text{s}$ and net displacement $=2\,\text{m}$. After $5$ cycles, time $=40\,\text{s}$ and displacement $=10\,\text{m}$. He then takes $3$ forward steps to reach $13\,\text{m}$, requiring $3\,\text{s}$. Total time $=40+3=43\,\text{s}$. The $x$-$t$ graph is a saw-tooth graph with slope $+1\,\text{m s}^{-1}$ for five seconds and $-1\,\text{m s}^{-1}$ for three seconds in each cycle.
$43\,\text{s}$.
$126\,\text{km h}^{-1}=35\,\text{m s}^{-1}$. Using $v^2=u^2+2as$ with $v=0$, $u=35\,\text{m s}^{-1}$, $s=200\,\text{m}$: $0=35^2+2a(200)$, so $a=-3.0625\,\text{m s}^{-2}$. Retardation is $3.1\,\text{m s}^{-2}$. Time $t=(v-u)/a=(0-35)/(-3.0625)=11.4\,\text{s}$.
Retardation $=3.1\,\text{m s}^{-2}$; stopping time $=11.4\,\text{s}$.
Acceleration due to gravity is always downward. At the highest point, instantaneous velocity is zero but acceleration remains $g$ downward. With downward as positive and origin at the highest point, the ball is below the origin during both upward and downward parts, so $x>0$; upward velocity is negative and downward velocity is positive; acceleration is positive. Maximum height $h=u^2/(2g)=29.4^2/(19.6)=44.1\,\text{m}$. Time to rise $=u/g=29.4/9.8=3.0\,\text{s}$, so return time $=6.0\,\text{s}$.
a. Downward; b. velocity $0$, acceleration $9.8\,\text{m s}^{-2}$ downward; c. $x>0$ and $a>0$ during both upward and downward motion, while $v<0$ upward and $v>0$ downward; d. height $44.1\,\text{m}$ and total time $6.0\,\text{s}$.
a. At the highest point of a vertically thrown ball, speed is zero but acceleration is $g$. b. Speed is the magnitude of velocity, so zero speed means zero velocity. c. In one-dimensional motion with reversals, speed can remain constant while velocity changes direction abruptly, so acceleration need not be zero at the reversal. d. If velocity is negative and acceleration positive, the particle slows down.
a. True; b. False; c. False; d. False.
First impact speed $v=\sqrt{2gh}=\sqrt{2(9.8)(90)}=42\,\text{m s}^{-1}$ and first fall time $t=\sqrt{2h/g}=4.29\,\text{s}$. After collision the speed becomes $0.9v=37.8\,\text{m s}^{-1}$. Time from floor to top is $37.8/9.8=3.86\,\text{s}$, so speed becomes zero at $t=8.14\,\text{s}$. It returns to the floor at $t=12.0\,\text{s}$ with speed $37.8\,\text{m s}^{-1}$.
The speed rises linearly from $0$ to $42\,\text{m s}^{-1}$ at $t\approx4.29\,\text{s}$, drops instantly to $37.8\,\text{m s}^{-1}$ after the bounce, falls linearly to $0$ at $t\approx8.14\,\text{s}$, and rises linearly to $37.8\,\text{m s}^{-1}$ at $t=12\,\text{s}$.
Displacement depends only on final and initial positions, while path length counts the actual distance travelled. If a person walks $5\,\text{m}$ forward and $3\,\text{m}$ backward, displacement magnitude is $2\,\text{m}$ but path length is $8\,\text{m}$. Dividing both quantities by the same time interval gives magnitude of average velocity and average speed respectively, so average speed is also greater than or equal to magnitude of average velocity. Equality holds only if the particle does not reverse direction in the interval.
Path length is always greater than or equal to magnitude of displacement. Average speed is always greater than or equal to magnitude of average velocity. Equality holds when motion is along one direction without reversal.
Time to market $=2.5/5=0.5\,\text{h}=30\,\text{min}$. i. In $30\,\text{min}$ he travels $2.5\,\text{km}$ with displacement $2.5\,\text{km}$, so both values are $2.5/0.5=5\,\text{km h}^{-1}$. ii. In $50\,\text{min}=5/6\,\text{h}$, he has returned home; displacement is zero and total path is $5\,\text{km}$, so average speed $=5/(5/6)=6\,\text{km h}^{-1}$. iii. In $40\,\text{min}=2/3\,\text{h}$, he has returned for $10\,\text{min}=1/6\,\text{h}$ and covers $7.5/6=1.25\,\text{km}$ back. Displacement magnitude $=2.5-1.25=1.25\,\text{km}$ and path length $=2.5+1.25=3.75\,\text{km}$, giving $1.25/(2/3)=1.875\,\text{km h}^{-1}$ and $3.75/(2/3)=5.625\,\text{km h}^{-1}$.
i. $|\bar v|=5\,\text{km h}^{-1}$, average speed $=5\,\text{km h}^{-1}$; ii. $|\bar v|=0$, average speed $=6\,\text{km h}^{-1}$; iii. $|\bar v|=1.875\,\text{km h}^{-1}$, average speed $=5.625\,\text{km h}^{-1}$.
Instantaneous velocity is the limiting value of displacement divided by time as the interval tends to zero. In that limit, the actual path segment is infinitesimal and has length equal to the magnitude of displacement. Hence instantaneous speed equals the magnitude of instantaneous velocity.
Because over an infinitesimal time interval, path length and displacement magnitude become the same.
In graph (a), one value of time corresponds to more than one position. In graph (b), one value of time corresponds to more than one velocity. In graph (c), speed becomes negative, which is impossible. In graph (d), total path length decreases with time, which is impossible because path length cannot decrease.
All four graphs cannot represent valid one-dimensional motion as labelled.
An $x$-$t$ graph shows position as a function of time, not the geometrical path in space. The graph can describe a particle at rest at $x=0$ for $t<0$, and then moving along a straight line with uniformly accelerated motion for $t>0$, so that $x\propto t^2$.
No.
Police van speed $=30\,\text{km h}^{-1}=8.33\,\text{m s}^{-1}$. Bullet speed relative to ground $=150+8.33=158.33\,\text{m s}^{-1}$. Thief's car speed $=192\,\text{km h}^{-1}=53.33\,\text{m s}^{-1}$. Speed of bullet relative to car $=158.33-53.33=105\,\text{m s}^{-1}$, which is the damaging speed.
$105\,\text{m s}^{-1}$.