Speed and Velocity Problems and Solutions 2

Q#8.

(II) A horse trots away from its trainer in a straight line, moving 38 m away in 9.0 s. It then turns abruptly and gallops halfway back in 1.8 s. Calculate

(a) its average speed and

(b) its average velocity for the entire trip, using “away from the trainer” as the positive direction.

Answer:

The distance traveled is

$38 \ m+\frac{1}{2}(38 \ m)=57 \ m$

and the displacement is

$38 \ m-\frac{1}{2}(38 \ m)=19 \ m$

The total time is 9.0 s + 1.8 s = 10.8 s

(a) $average \ speed = \frac{distance}{time \ elapsed}$

$average \ speed = \frac{57 \ m}{10.8 \ s}=5.3 \ m/s$

(b) $average \ velocity = v_{avg} = \frac{displacement}{time \ elapsed}$

$v_{avg}= \frac{19 \ m}{10.8 \ s}=5.3 \ m/s$

Q#9

(II) A person jogs eight complete laps around a 400-m track in a total time of 14.5 min. Calculate

(a) the average speed and

(b) the average velocity, in m/s.

Answer:

The distance traveled is 3200 m (8 laps × 400 m/lap).

That distance probably has either 3 or 4 significant figures, since the track distance is probably known to at least the nearest meter for competition purposes.

The displacement is 0, because the ending point is the same as the starting point.

(a) $average \ speed = \frac{d}{\Delta t}$

$average \ speed = \frac{3200 \ m}{14.5 \ min} \times \frac{1 \ min}{60 \ s}$

$average \ speed = 3.68 \ m/s$

$average \ velocity = \frac{\Delta x}{\Delta t}0 \ m/s$

Q# 10

(II) Every year the Earth travels about as it orbits the Sun. What is Earth’s average speed in km/h.

Answer:

The average speed is the distance divided by the time.

$average \ speed=\frac{d}{ t}$

$average \ speed=\left(\frac{1 \times 10^9 \ km}{ 1 \ yr}\right)\left(\frac{1 \ yr}{365.25 \ d}\right)\left(\frac{1 \ d}{24 \ h}\right)$

$average \ speed=1.145 \times 10^5 \ km/h \approx 1 \times 10^5 \ km/h$

Q# 11

(II) A car traveling is 210 m behind a truck traveling 75 km/h. How long will it take the car to reach the truck?

Answer:

Both objects will have the same time of travel. If the truck travels a distance $d_{truck}$ , then the distance the car travels will be $d_{car}=d_{truck}+210 \ m$ .

Using the equation for average speed, $\overline{v}=\frac{d}{\Delta t}$ , solve for time, and equate the two times.

Read More:

$\Delta t = \frac{d_{truck}}{\overline{v}_{truck}}=\frac{d_{car}}{\overline{v}_{car}}$

$\frac{d_{truck}}{75 \ km/h}=\frac{d_{truck}+210 \ m}{95 \ km/h}$

$(75 \ km/h)(d_{truck})+(75 \ km/h)(210 \ m)=(95 \ km/h)d_{truck}$

$d_{truck}=(210 \ m)\frac{75 \ km/h}{95 \ km/h - 75 \ km/h}=787.5 \ m$

The time of travel is

$\Delta t =\frac{d_{truck}}{\overline{v}_{truck}}$

$\Delta t =\frac{787.5 \ m}{75,000 \ m/h}\times \frac{60 \ min}{h}$

$\Delta t = 0.63 \ min = 37.8 \ s \approx 38 \ s$

Also note that

$\Delta t =\frac{d_{car}}{\overline{v}_{car}}$

$\Delta t =\frac{787.5 \ m+210 \ m}{95,000 \ m/h}\times \frac{60 \ min}{h}$

$\Delta t = 0.63 \ min = 37.8 \ s \approx 38 \ s$

ALTERNATE SOLUTION:

The speed of the car relative to the truck is 95 km/h − 75 km/h = 20 km/h.

In the reference frame of the truck, the car must travel 210 m to catch it.

$\Delta t =\frac{0.21 \ km}{20 \ m/h}\times \frac{3600 \ s}{1 \ h}=37.8 \ s$

Q#12.

(II) Calculate the average speed and average velocity of a complete round trip in which the outgoing 250 km is covered at 95 km/h followed by a 1.0-h lunch break, and the return 250 km is covered at 55 km/h.

Answer:

The distance traveled is 500 km (250 km outgoing, 250 km return, keep 2 significant figures).

The displacement (Δx) is 0 because the ending point is the same as the starting point.

To find the average speed, we need the distance traveled (500 km) and the total time elapsed.

During the outgoing portion,

$\overline{v}_1=\frac{\Delta x_1}{\Delta t_1}$ , so