Question

A partical travels one fourth of the dist with constant speed of 5 m/s and remaining dist. with thr constant speed of 18 m/s . What is the average speed of the car ?

Answer 1

Explanation:

total distance=s

T1= s/4 ÷ 5

=s/20

T2=3s/4÷ 18

= s/24

average speed=s/ T1+T2

=s/ s/20+s/24

.....take s as common and cancel it..

1/1/20+1/24

1/ 6+5/120

120/11

10.9m/s

Answer 2

Given: Distance = d

           d/4 distance covered with speed = 5m/s

            remaining distance  covered with speed = 18m/s

To Find: average speed of the car - v =?

Solution:

Formula used:

• Average speed = Total distance covered (d) /total time taken(T)
• Distance = speed × time

d/4 distance, speed = 5m/s  (journey I)

d - d/4 = $\frac{4d - d}{4}$ = 3d/4 distance speed = 18m/s  (journey II)

Total distance = d

1. Calculating time take for Journey I = t₁

Time  = Distance/speed (from the above Formula)

t₁ = $\frac{\frac{d}{4} }{5}$  

t₁ = $\frac{d}{4 * 5}$

t₁ = d/20  

    2. Calculating time taken for Journey II = t₂

t₂ = $\frac{\frac{3d}{4} }{18}$

t₂ = $\frac{3d}{4 * 18}$

t₂ = $\frac{3d}{72}$

t₂ = d/24

    3. Calculating Total time taken = T

T = d/20 + d/24

T = $\frac{6d + 5d}{120}$

T = 11d/120

    4. Calculating Average velocity = v

v = d/T

v = $\frac{d}{\frac{11d}{120} }$

v = $\frac{d}{11d} * 120$

v = 120/11

v = 10.9 m/s

Hence, the average speed of the car = 10.9 m/s