Question

1. $s = 600\\,\text{m} / 15\\,\text{s}$\

\underline{40m/s} \

1. $s = 240\\,\text{km} / 4\\,\text{hr}$\

\underline{60km/hr} \

1. $s = 75\\,\text{mi} / 2.5\\,\text{hr}$\

\underline{30mi/hr} \

1. $s = (35\\,\text{m} - 20\\,\text{m}) / (10\\,\text{s} - 5\\,\text{s})$\

\underline{15m/5s = 3m/s} \

1. $s = (46\\,\text{m} - 18\\,\text{m}) / (10\\,\text{s} - 3\\,\text{s})$\

\underline{28m/7s = 4m/s} \

1. $s = (49\\,\text{m} - 21\\,\text{m}) / (14\\,\text{s} - 7\\,\text{s})$\

\underline{28m/7s = 4m/s} \
\
use the graph or the formula for speed to answer questions 7–12.\
\

1. a trucker made a delivery to a town 180 km from his start point. the graph shows the time and distance for the trip. during which part of the trip was the trucker driving 55 km/h?\

\underline{150km - 40km = 110km; 3hr - 1hr = 2hr; 110km/2hr = 55km/hr} \

1. the trucker stopped at a truck stop for a one - hour lunch break. during which part of the trip did he take his lunch break?\

\underline{140km - 140km = 0km; 4hr - 3hr = 1hr; 0km/1hr = 0km/hr} \

1. what was the trucker’s speed as he drove from the truck stop where he had lunch to his final destination?\

\underline{180km - 140km = 40km; 5hr - 4hr = 1hr; 40km/1hr = 40km/hr} \

1. a jogger runs along a road for a distance of 2700 m. if it takes her 900 seconds to run that distance, what is her speed?\

\underline{$v = \frac{s}{t}$; $\frac{2700\\,\text{m}}{900\\,\text{s}} = 3\\,\text{m/s}$} \

1. a car travels 40 miles in the first hour and 50 miles in the second hour. what is the car’s average speed over the entire trip?\

\underline{$\frac{50\\,\text{mi} - 40\\,\text{mi}}{2\\,\text{hr} - 1\\,\text{hr}} = \frac{10\\,\text{mi}}{1\\,\text{hr}} = 10\\,\text{mi/hr}$} \

1. a bicyclist travels 10 km in half an hour, then rests for half an hour, then travels 50 km in three hours. what was the bicyclist’s average speed over the entire trip?\

\
(graph: distance - time record for trip a, with time (hours) on x - axis (0 - 5) and distance (km) on y - axis (0 - 180))\
\
motion and forces, chapter 1, math practice 51

Explanation:

Question 10

Step1: Recall the speed formula

The formula for speed \( S \) is \( S=\frac{d}{t} \), where \( d \) is distance and \( t \) is time.

Step2: Substitute the given values

We know \( d = 2700\space m \) and \( t=900\space s \). Substitute these into the formula: \( S=\frac{2700}{900} \).

Step3: Calculate the result

\( \frac{2700}{900}=3 \).

Step1: Find total distance

The car travels 40 miles in the first hour and 50 miles in the second hour. So total distance \( d = 40 + 50=90\space miles \).

Step2: Find total time

Total time \( t = 1+ 1 = 2\space hours \).

Step3: Use speed formula

Using \( S=\frac{d}{t} \), substitute \( d = 90 \) and \( t = 2 \): \( S=\frac{90}{2} \).

Step4: Calculate the result

\( \frac{90}{2}=45 \).

Step1: Find total distance

The bicyclist travels 10 km and then 50 km. So total distance \( d=10 + 50 = 60\space km \).

Step2: Find total time

The times are \( 0.5\space hr \) (traveling), \( 0.5\space hr \) (resting), and \( 3\space hr \) (traveling). Total time \( t=0.5 + 0.5+3=4\space hours \).

Step3: Use speed formula

Using \( S=\frac{d}{t} \), substitute \( d = 60 \) and \( t = 4 \): \( S=\frac{60}{4} \).

Step4: Calculate the result

\( \frac{60}{4}=15 \).

Answer:

The jogger's speed is \( 3\space m/s \).

Question 11