Question

1,
s the first 5 km of its journey at a
113. A train covers the first 5 km of
speed of 30 km/hr and the near
speed of 45 km/hr. What is the
of the train ?
ar and the next 15 km at a
km/hr. What is the average speed
(a)
35 km/hr
(b)
37.5 km/hr
(c)
39-5 km/hr
(d)
40 km/hr​

Answer 1

$\huge\underline\blue{\sf Answer}$

$\large\red{\boxed{\sf V_{av}=40\:km/hr}}$

$\huge\underline\blue{\sf Solution}$

$\large\underline\pink{\sf Given:}$

• Train covers first distance $\sf{s_1}$= 5 km

• Covers first distance with speed $\sf{v_1}$=30 km/hr

• Train covers second distance $\sf{s_2}$=15 km

• Covers second distance with the speed of $\sf{v_2}$=45 km/hr

$\large\underline\pink{\sf To\:Find:}$

• Average speed of train $\sf{V_{av}}$ = ?

_______________________________

We know that ,

$\large{\boxed{\sf Speed(v)={\frac{Distance(s)}{Time(t)}}}}$

Now,

For first distance :

$\large\implies{\sf t_1={\frac{5}{30}}}$

$\large\implies{\sf t_1={\frac{1}{6}}hr}$

For second distance :

$\large\implies{\sf t_2={\frac{15}{45}}}$

$\large\implies{\sf t_2={\frac{1}{3}}hr}$

________________________________

As we know ,

$\large{♡}$$\large{\boxed{\sf V_{av}={\frac{Total\:Distance}{Total\:Time}}}}$

$\large\implies{\sf {\frac{s_1+s_2}{t_1+t_2}}}$

On putting value :

$\large\implies{\sf {\frac{5+15}{1/6+1/3}}}$

$\large\implies{\sf {\frac{20}{3+6/6 \times 3}}}$

$\large\implies{\sf {\frac{20 \times 18}{9}}}$

$\large\implies{\sf V_{av}=40\:km/hr}$

$\huge\red{♡}$$\large\red{\boxed{\sf V_{av}=40\:km/hr}}$