Question

Question :-

An Object Travels 24m with speed of 4 m\s and then another 32m with the speed of 8 m\s. The average speed of the object is

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Answer 1

Answer:

D-Distance, S-Speed, T-Time.

D1 = 24m

S1 = 4m/s

:. T1 = D1/S1 = 24/4 = 6s

D2 = 32m

S2 = 8m/s

:. T2 = D2/S2 = 32/8 = 4s

Avg. S = Total D / Total T

= (24+32)/(6+4)

= 56/10

= 5.6 m/s

Explanation:

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Answer 2

Given :

• An Object Travels 24 m with speed of 4 m\s .
• Other object travels 32 m with the speed of 8 m\s.

To Find :

• Average Speed .

Solution :

Firstly we'll find the Time taken by both objects :

For 1st Object :

$\longmapsto\tt\boxed{Time=\dfrac{Distance}{Speed}}$

$\longmapsto\tt{t=\dfrac{24}{4}}$

$\longmapsto\tt\bf{t=6\:sec}$

For 2nd Object :

$\longmapsto\tt\boxed{Time=\dfrac{Distance}{Speed}}$

$\longmapsto\tt{t=\dfrac{32}{8}}$

$\longmapsto\tt\bf{t=4\:sec}$

Now ,

$\longmapsto\tt{{Distance}_{1}=24\:m}$

$\longmapsto\tt{{Distance}_{2}=32\:m}$

$\longmapsto\tt{{Time}_{1}=6\:sec}$

$\longmapsto\tt{{Time}_{2}=4\:sec}$

Using Formula :

$\longmapsto\tt\boxed{Average\:Speed=\dfrac{Total\:Distance}{Total\:Time\:Taken}}$

Putting Values :

$\longmapsto\tt{\dfrac{24+32}{6+4}}$

$\longmapsto\tt{\dfrac{24+32}{6+4}}$

$\longmapsto\tt{\dfrac{56}{10}}$

$\longmapsto\tt\bf{5.6\:m/s}$

So , The Average Speed is 5.6 m/s .