Question

A car moves with a speed of 30 km/hr-1 for half an hour, 25 km/hr for one hour and 40 km/hr-1 for two hours. Calculate the average speed of the car

Answer 1

Answer:

The Average Speed is 34.28 km/hr.

Explanation:

Speed 1 = 30 km/hr

Time 1 = 0.5 hr

Distance = Speed * Time

= [30*0.5] km

= 15 km

Speed 2 = 25 km/hr

Time 2 = 1 hr

D = S*T

= [25*1] km

= 25 km

Speed 3 = 40 km/hr

Time 3 = 2 hrs

D = S*T

= [40*2] km

= 80 km

Average Speed = Total Distance Covered/Total Time Taken

= (15+25+80) / (0.5+1+2)

= [120/3.5] km/hr

= 34.28 km/hr

Hope this helps.

Answer 2

$\large\underline{\bigstar \: \: {\sf Given-}}$

• Car moves with a speed of ${\sf (v_1)=30\:km/h \: (v_2)=25\:km/h\: ,(v_3)=40\: km/h }$
• Time ${\sf (t_1)=0.5hr, \: (t_2)=1hr, \: (t_3)=2\:hr}$

$\large\underline{\bigstar \: \: {\sf To \: Find -}}$

• Average speed of car ${\sf (v_{av})}$

$\large\underline{\bigstar \: \: {\sf Formula \: Used -}}$

$\bullet\underline{\boxed{\sf Average \:Speed v_{av}=\dfrac{Total \: Distance}{Total \: Time}}}$

$\large\underline{\bigstar \: \: {\sf Solution-}}$

➩ Displacement (s)

$\implies\boxed{\sf Distance (s)=speed\times time }$

$\implies{\sf s_1=v_1t_1}$

$\implies{\sf s_1=30\times 0.5}$

$\implies{\bf s_1=15\:km}$

$\implies{\sf s_2=v_2t_2 }$

$\implies{\sf s_2=25\times 1}$

$\implies{\bf s_2=25\:km}$

$\implies{\sf s_3=v_3t_3}$

$\implies{\sf s_3=40\times 2 }$

$\implies{\bf s_3=80\:km}$

$\implies{\sf s = s_1+s_2+s_3 }$

$\implies{\sf s= 15+25+80}$

$\implies{\bf s=120\:km}$

_________________________________

$\implies{\sf Average \: Speed (v_{av})=\dfrac{Distance}{Time} }$

$\implies{\sf \dfrac{120}{0.5+1+2}}$

$\implies{\sf \dfrac{120}{3.5} }$

$\implies{\bf v_{av}=34.2\: km/h}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀OR

$\implies{\sf v_{av}=34.2\times \dfrac{5}{18}}$

$\implies{\bf v_{av}=9.52\: m/s}$

$\large\underline{\bigstar \: \: {\sf Answer-}}$

Average speed of car is ${\bf 34.2\: km/h \; or \: 9.52\: m/s}$