Question

4. A man travels one third length of a straight road with a constant speed
of 40 m/sec. and the remaining portion of the road with a constant
speed of 60 m/sec., then calculate its average speed.
Ans. 51.4 m/sec.​

Answer 1

✵ Explanation

Total Distance=x

S1= 60 m/ s.D2=2x/3.t2= 2x/3×60

Average speed= total distance traveled÷ total time

avg.speed = x/3x+4x/360

avg.speed = 360/7 =51.4 m/s......

Answer 2

Let the total distance covered by the man be d

Part I :

Distance traveled = d/3

Speed  of the man = 40 m/s

we know that ,

$\implies\mathtt{speed =\dfrac{distance}{time}}$

$\implies\mathtt{time =\dfrac{distance}{speed}}$

$\implies\mathtt{t_1=\dfrac{d}{120}}$

----------------------------

Part II:

Remaining distance = d - d/3 =2d/3

Speed of the man = 60 m/s

we know that,

$\implies\mathtt{time =\dfrac{distance}{speed}}$

$\implies\mathtt{t_2 =\dfrac{2d}{180}}$

$\implies\mathtt{t_2 =\dfrac{d}{90}}$

---------------------------

Average speed is given by the formula,

$\implies\mathtt{Average\: speed = \dfrac{total\: distance}{total \:time} }$

$\implies\mathtt{Average\: speed = \dfrac{d}{t_1+t_2} }$

$\implies\mathtt{Average\: speed = \dfrac{d}{\dfrac{d}{120} +\dfrac{d}{90}} }$

$\implies\mathtt{Average\: speed = \dfrac{d}{\dfrac{3d+4d}{360} }}$

$\implies\mathtt{Average\: speed = \dfrac{d\times360}{7d}}$

$\implies\mathtt{Average\: speed = \dfrac{360}{7}}$

$\red{\implies\mathtt{Average\: speed =51.4 \dfrac{m}{s}}}$

Additional Formulas:

First equation of motion,

$\implies\mathtt{v= u+ at}$

Second equation of motion,

$\implies\mathtt{s= ut+\dfrac{1}{2} at^{2} }$

Third equation of motion ,

$\implies\mathtt{v^{2} = u^{2} +2 as}$

To find distance at n' th second ,

$\implies\mathtt{s_n= u + \dfrac{a}{2} (2n-1)}$

here ,

• v= final velocity
• u = initial velocity
• t = time taken
• s = distance covered
• a = acceleration