Question

A car covers the first half of the distance between two station at a speed of 60km/h and second half at 80 km/h. calculate the average speed of the car.

Answer 1

Answer:Letx consider, x= total distance covered by the car,

Time taken to complete the half distance with velocity v=40km/h,

t  

1

​  

=  

2×40

x

​  

=  

80

x

​  

hr. . . . .(1)

The time taken to complete another half distance with velocity u=60km/h

t  

2

​  

=  

2×60

x

​  

=  

120

x

​  

hr

The total time taken, t=t  

1

​  

+t  

2

​  

 

t=  

80

x

​  

+  

120

x

​  

=  

48

x

​  

hr

The average speed of the car,  

v  

avg

​  

=  

x/48

x

​  

=48km/h

Explanation:

Answer 2

Answer:

Let the total distance covered be s.

Given -

• Speed of the car in the first half of distance, $\sf{t}_{1}$ = 60 km/h

Let's find the time taken by car in first half:

$\\ \sf \: t_1 = \frac{{}^{s} / \: {2}^{}}{60} \\ \\ \\ \implies \large{ \boxed{ \sf \: t_1 = \dfrac{s}{120} }}$

• Speed of the car at the second half of the distance, $\sf{t}_{2}$ = 80 km/h

Time taken by car in second half :

$\\ \sf \: t_2 = \frac{ {}^{s}/ \: {2}^{} }{80 } \\ \\ \\ \implies \large{ \boxed{ \sf{t_2 = \dfrac{s}{160} }}}$

$\\ \: \large{ \underline{ \mathfrak{ \: \star \: Average \: \: speed \: - }}} \\ \\ \\ \implies \sf \: \frac{total \: \: distance \: \: travelled}{total \: \: time \: \: taken} \\ \\ \\ \implies \sf \: \frac{s}{ \dfrac{s}{120} + \dfrac{s}{160} } \\ \\ \\ \implies \sf \: \frac{120 \times 160}{120 + 160} \\ \\ \\ \implies \sf \: \dfrac{19200}{280} \\ \\ \\ \implies \sf {\underline{68.5 \: kmph}}$

Therefore, the average speed of the car is 68.5 km/h.