Q#1
If you are driving 95 km/h along a straight road and you look to the side for 2.0 s, how far do you travel during this inattentive period?
Answer:
Given:
driving speed, v = 95 km/h = $95 \frac{km}{h}\times \frac{1000 \ m/km}{3600 \ s/h}=26.4 \ m/s$
driving time, t = 2,0 s
The distance of travel (displacement) can be found by rearranging Eq. $\overline{v}=\frac{\Delta x}{\Delta t}$ for the average velocity.
Also note that the units of the velocity and the time are not the same, so the speed units will be converted.
$\Delta x = \overline{v}.\Delta t$
$\Delta x = (26.4 \ m/s)(2.0 \ s)=52.8 \ m$ = 0.0528 km
Q#2
What must your car’s average speed be in order to travel 235 km in 2.75 h?
Answer:
Gives:
distance of car's, d = 235 km = (235 km)(1000 m/km) = 235,000 m
time, $\Delta t$ = 2.75 h = 9,900 s
The average speed is given by Eq. $\overline{v}=\frac{d}{\Delta t}$, using d to represent distance traveled.
$\overline{v}=\frac{235 \ km}{2.75 \ h}=85.5 \ km/h$ or
$\overline{v}=\frac{235,000 \ m}{9,900 \ s}=23.7 \ m/s$
Q#3
A particle at $t_1=-2.0 \ s$ is at $x_1=4.8 \ cm$ and at $t_2=4.5 \ s$ is at $x_2=8.5 \ cm$. What is its average velocity over this time interval? Can you calculate its average speed from these data?Why or why not?
Answer:
The average velocity is given by Eq.
$\overline{v}=\frac{\Delta x}{\Delta t}$
$\overline{v}=\frac{x_2-x_1}{t_2-t_1}$, so
$\overline{v}=\frac{8.5 \ cm-4.8 \ cm}{4.5 \ s-(-2.0 \ s)}$
$\overline{v}=0.57 \ cm/s$
The average speed cannot be calculated. To calculate the average speed, we would need to know the actual distance traveled, and it is not given. We only have the displacement.
Q#4
A rolling ball moves from $x_1=8.4 \ cm$ to $x_2=-4.2 \ cm$ during the time from $t_1=3.0 \ s$ to $t_2 = 6.1 \ s$. What is its average velocity over this time interval?
Answer:
The average velocity is given by Eq.
$\overline{v}=\frac{\Delta x}{\Delta t}$
$\overline{v}=\frac{x_2-x_1}{t_2-t_1}$, so
$\overline{v}=\frac{-4.5 \ cm-8.4 \ cm}{6.1 \ s-3.0 \ s}$
$\overline{v}=\frac{-12.6 \ cm}{3.1 \ s}=-4.1 \ cm/s$
The negative sign indicates the direction.
Q#5
A bird can fly 25 km/h. How long does it take to fly 3.5 km?
Answer:
The time of travel can be found by rearranging the average velocity equation.
$\overline{v}=\frac{\Delta x}{\Delta t}$, so
$\Delta t = \frac{\Delta x}{\overline{v}}$
$\Delta t = \frac{3.5 \ km}{25 \ km/h}$
$\Delta t = 0.14 \ h = 8.4 \ min = 504 \ s$
Q#6
According to a rule-of-thumb, each five seconds between a lightning flash and the following thunder gives the distance to the flash in miles.
(a) Assuming that the flash of light arrives in essentially no time at all, estimate the speed of sound in m/s from this rule.
(b) What would be the rule for kilometers?
Answer:
(a) The speed of sound is intimated in the problem as 1 mile per 5 seconds. The speed is calculated as follows:
$speed=\frac{distance}{time}$
$speed=\left(\frac{1 \ mi}{5 \ s}\right)\left(\frac{1610 \ m}{1 \ mi}\right)$
$speed=322 \ m/s \approx 300 \ m/s$
(b) The speed of 322 m/s would imply the sound traveling a distance of 966 meters (which is approximately 1 km) in 3 seconds.
So the rule could be approximated as 1 km every 3 seconds.
Q#7
You are driving home from school steadily at 95 km/h for 180 km. It then begins to rain and you slow to 65 km/h. You arrive home after driving 4.5 h.
(a) How far is your hometown from school?
(b) What was your average speed?
Answer:
(a) The average speed is given by Eq.
$\overline{v}=\frac{d}{\Delta t}$
The time for the first part of the trip is calculated from the initial speed and the first distance, using d to represent distance.
$\overline{v_1}=\frac{d_1}{\Delta t_1}$
$\Delta t_1=\frac{d_1}{\overline{v_1}}=\frac{180 \ km}{95 \ km/h}$
$\Delta t_1=1.895 h = 113.7 \ min$
The time for the second part of the trip is now calculated.
$\Delta t_2=\Delta t_{total}-\Delta t_1$
$\Delta t_2=4.5 \ h = 1.895 \ h=2.605 \ h=156.3 \ min$
The distance for the second part of the trip is calculated from the average speed for that part of the trip and the time for that part of the trip.
$\overline{v_2}=\frac{d_2}{\Delta t_2}$
$d_2=\overline{v_2}.\Delta t_2$
$d_2=(65 \ km/h)(2.605 \ h)=169.3 \ km \approx 170 \ km$
So, the total distance is then
$d_{total}=d_1+d_2$
$d_{total}$ = 180 km + 169.3 km = 349.3 km $\approx$ 350 km
(b) The average speed is NOT the average of the two speeds. Use the definition of average speed, Eq. $\overline{v}=\frac{d_{total}}{\Delta t_{total}}$
$\overline{v}=\frac{349.3 \ km}{4.5 \ h}$
$\overline{v}=77.62 \ km/h \approx 78 \ km/h$
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