Question

- A student starts to the school at a speed 18 km/hr after 10 minutes, he thought he will be late
for school so he increased speed to 27 km/hr and reached school in 5 minutes. What is
average speed of student?​

Answer 1

Gɪᴠᴇɴ :-

• Initial Speed of Student = 18km/h .
• Travel for = 10 Minutes.
• Increased Speed to = 27km/h.
• Now Reached in Next = 5 Minutes .

Tᴏ Fɪɴᴅ :-

• Average Speed of Student ?

Fᴏʀᴍᴜʟᴀ ᴜsᴇᴅ :-

• Distance = Speed * Time.
• Time = ( Distance ) / Speed .
• Speed = (Distance) / Time.
• Average Speed = (Total Distance) / (Total Time Taken) .

Sᴏʟᴜᴛɪᴏɴ :-

Let us Assume That, Distance from student school to his Home is x km.

Case 1 :-

→ Speed = 18km/h .

→ Time = 10 min = (10/60) = (1/6) Hours.

→ Distance covered by Student = S * T = 18 * (1/6) = 3km.

Now,

Case 2 :-

→ New Speed inc. to = 27km/h.

→ Time to Cover Rest Distance = 5 min. = (5/60) = (1/12) Hours.

→ Distance = S * T = 27 * (1/12) = 2.25km .

Therefore,

→ Total Distance from Home to school = x = 3 + 2.25 = 5.25 km.

Now,

→ Total Time Taken By student = 10min + 5 min = 15 min = (15/60) = (1/4) Hours.

Hence,

→ Average Speed = (5.25) / (1/4) = 5.25 * 4 = 21 km/h (Ans.)

∴ Average Speed of Student is 21km/h.

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

• Intial Speed = 18km/h
• Time =10min
• Increased speed = 27km/hr
• Reached school in =5 min

${\bf{\blue{\underline{To\: Find:}}}}$

• Average speed =?

${\bf{\blue{\underline{Now:}}}}$

When he reached at school in 10 min,

• Intial speed of student = 18km\hr

$\star{\bf{Time \: taken \: = 10min }}$

$\implies{\bf{ \frac{10}{60} }}$

$\implies{\bf{ \frac{1}{6} }}$

$\star\:{\bf{Distance = S \times t }}$

$\implies{\bf{ 18 \times \frac{1}{6} }}$

$\implies{\bf{3 }}$

When he reached at school in 5min,

$\star{\bf{Speed \: of \: student \: = 27km \: h }}$

$\star{\bf{Time \: taken = 5 \: min }}$

$\implies{\bf{ \frac{5}{60} }}$

$\implies{\bf{ \frac{1}{12} }}$

$\star{\bf{distance \: = S \times T }}$

$\implies{\bf{27 \times \frac{1}{12} }}$

$\implies{\bf{2.25km }}$

$\star{\bf{ \purple{ Total \: Distance \: = 3 + 2.25}}}$

$\star{\bf{ \purple{ Total \: Distance \: = 5.25}}}$

$\star{\bf{Total \: Time = 10 + 5 \: min }}$

$\implies{\bf{ 15 \: min }}$

$\implies{\bf{ \frac{15}{60} hr }}$

$\implies{\bf{ \frac{1}{4} hr }}$

$\boxed{\bf{ \purple{ Average \: Speed \: = \frac{Total \: Distance}{Total \: Time} }}}$

$\implies{\bf{ \frac{5.25}{ \frac{1}{4} } }}$

$\implies{\bf{ 5.25 \times 4 }}$

$\implies{\bf{ 21km hr }}$

${\bf{\orange{\underline{Speed:}}}}$

• Speed of a particular can either be zero or positive

${\bf{\orange{\underline{Average\: Speed:}}}}$

• A average speed of particle in a given interval of time is given by the ratio of the total distance travelled by it to Total time taken.