Question

9. The odometer of a car reads 4000 km at hest
of a trip and 4600 km at the end of the
trip took 10 h. calculate the average speed of the
car in a. km/h and b. m/s. Ans.​

Answer 1

Answer:

$distance = 4600 - 4000 = 600 km\\ now \: \\ \: \: \: \: \: \: \ time = 10hrs \\ speed = distance | time \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: a. 600 \div 10 = 60km | hr \\ b.60000 \div 60 \times 60 \times 24 \\ = 0.694$

so the answers are a.60km/he

B. 0.694m/min

Answer 2

Answer:

• The average speed (v) in Km/h is 60 Km/h
• The average speed (v) in m/s is 16.67 m/s

Given:

1. Initial distance = 4000 Km
2. Final Distance = 4600 Km
3. Time taken (t) = 10 hours

Explanation:

$\rule{300}{1.5}$

From the formula we know,

$\displaystyle \dashrightarrow\sf Average \; Speed=\dfrac{Total\;Distance}{Total\;Time}$

Now,

$\displaystyle\dashrightarrow\sf v=\dfrac{Final\;Distance\;-\;Initial\;Distance}{Total \;time\;taken}$

Substituting the values,

$\displaystyle \dashrightarrow\sf v=\dfrac{4600\;-\;4000}{10}\\\\\\\dashrightarrow\sf v=\dfrac{600}{10}\\\\\\\dashrightarrow\sf v=\cancel{\dfrac{600}{10}}\\\\\\\dashrightarrow\sf v = 60\\\\\\\dashrightarrow \large{\underline{\boxed{\red{\sf v=60\;Km/h }}}}$

∴ The average speed (v) in Km/h is 60 Km/h.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From the conversion unit we know,

$\displaystyle\dashrightarrow\sf 1 Km/h=\dfrac{5}{18}\;m/s$

Substituting and solving,

$\displaystyle\dashrightarrow\sf 1\;Km/h = \dfrac{5}{18}\;m/s\\\\\\\dashrightarrow\sf 60\; Km/h=60\times\dfrac{5}{18}\\\\\\\dashrightarrow\sf 60\; Km/h=\dfrac{300}{18}\\\\\\\dashrightarrow\sf 60\; Km/h=\cancel{\dfrac{300}{18}}\\\\\\\dashrightarrow\sf 60\;Km/h=16.67\\\\\\\dashrightarrow \large{\underline{\boxed{\red{\sf v=16.67\;m/s }}}}$

∴ The average speed (v) in m/s is 16.67 m/s.

$\rule{300}{1.5}$