Question

if a truck travels 170 kilometers north for 3 hours and then it goes to travel and covered 150 kilometers going to south for 2.5 hours.
a. What is its average speed in km/hr and in m/s?
b. And what is its average velocity in km/hr and in m/s?

Answer 1

Average Speed:

$avg. \: u = \dfrac{total \: distance}{total \: time}$

$\implies avg. \: u = \dfrac{170 + 150}{3 + 2.5}$

$\implies avg. \: u = \dfrac{320}{5.5}$

$\boxed{ \implies avg. \: u = 58.18 \: km/hr}$

$\implies avg. \: u = 58.18 \times \dfrac{5}{18}$

$\boxed{\implies avg. \: u = 16.1 \: m/s}$

Average Velocity:

$avg. \: v = \dfrac{total \: displacement}{total \: time}$

$\implies avg. \: v = \dfrac{170 - 150}{3 + 2.5}$

$\implies avg. \: v = \dfrac{20}{5.5}$

$\boxed{ \implies avg. \: v = 3.63 \: km/hr}$

$\implies avg. \: v = 3.63 \times \dfrac{5}{18}$

$\boxed{\implies avg. \: v = 1.08 \: m/s}$

Hope It Helps.

Answer 2

Answer:

Average Speed:

avg. \: u = \dfrac{total \: distance}{total \: time}avg.u=

totaltime

totaldistance

\implies avg. \: u = \dfrac{170 + 150}{3 + 2.5}⟹avg.u=

3+2.5

170+150

\implies avg. \: u = \dfrac{320}{5.5}⟹avg.u=

5.5

320

\boxed{ \implies avg. \: u = 58.18 \: km/hr}

⟹avg.u=58.18km/hr

\implies avg. \: u = 58.18 \times \dfrac{5}{18}⟹avg.u=58.18×

18

5

\boxed{\implies avg. \: u = 16.1 \: m/s}

⟹avg.u=16.1m/s

Average Velocity:

avg. \: v = \dfrac{total \: displacement}{total \: time}avg.v=

totaltime

totaldisplacement

\implies avg. \: v = \dfrac{170 - 150}{3 + 2.5}⟹avg.v=

3+2.5

170−150

\implies avg. \: v = \dfrac{20}{5.5}⟹avg.v=

5.5

20

\boxed{ \implies avg. \: v = 3.63 \: km/hr}

⟹avg.v=3.63km/hr

\implies avg. \: v = 3.63 \times \dfrac{5}{18}⟹avg.v=3.63×

18

5

\boxed{\implies avg. \: v = 1.08 \: m/s}

⟹avg.v=1.08m/s

Explanation:

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