Question

A particle moving in a straight line covers half of the total distance with a speed 3m/s. The other half of the distance is covered in two equal time intervals with speed 4.5m/s and 7.5m/s. The average speed of the particle during this motion is?
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Answer 1

Given :-

• Speed for the first half of the distance = 3m/s
• Speed 2 and 3 have the same time interval
• The second half of the distance, the object travels in 2 speeds : 4.5m/s and 7.5m/s respectively

Aim :-

• To calculate the average speed of the object

Answer :-

Formula to use :

$\boxed{\sf Speed = \dfrac{Distance}{Time} }$

$\boxed {\sf Speed \times Time = Distance}$

$\boxed {\sf Average\:Speed = \dfrac{Total\:distance\:travelled}{Total\:time\:taken} }$

Solution :

Speed 1 :

Distance = d/2

Time = T

Speed = 3m/s

By substituting these values in the equation speed = Distance/Time,

$\implies \sf 3 = \dfrac{d}{\dfrac{2}{T} }$

$\implies \sf 3 = \dfrac{d}{2} \div T$

$\implies \sf 3 = \dfrac{d}{2} \times \dfrac{1}{T}$

$\implies \sf 3 = \dfrac{d}{2T}$

Transposing 2T,

$\implies \sf 3\times 2T = d$

$\implies \sf 6T = d$

Speed 2 :

Distance = D₂

Time = t

Speed = 4.5m/s

Substituting these values in the equation,

$\implies \sf 4.5 = \dfrac{D_{2}}{t}$

Transposing t to the other side,

$\implies \sf 4.5 \times t = D_{2}$

$\implies \sf 4.5t = D_{2}$

Speed 3 :

Distance = D₃

Time = t

Speed = 7.5 m/s

Substituting the value in the equation,

$\implies \sf 7.5 = \dfrac{D_{3}}{t}$

Transposing t  to the other side,

$\implies \sf 7.5 \times t = D_{3}$

$\implies \sf 7.5t = D_{3}$

In the question, it's given that in the first half of the duration, the speed was 3m/s and in the second half it was 4.5m/s and 7.5m/s

Hence,

d/2 = D₂ + D₃

By solving the equations above we obtained D₂ and D₃ to be 4.5t and 7.5t respectively.

By adding both of the distances we get :

⇒ 4.5t + 7.5t

⇒ 12t

Hence d/2 = 12t

After deriving Total distance to be 6T from the first equation, d/2 = 6T/2.

Equating,

$\implies \sf \dfrac{6T}{2} = 12t$

Transposing 2,

$\implies \sf 6T = 12t \times 2$

$\implies \sf 6T = 24t$

Transposing 6,

$\implies \sf T = \dfrac{24t}{6}$

$\implies \sf T = 4t$

This implies that the total distance is 24t and value of T is 4t.

We now know that the total distance = 24t

The total time will be :

⇒ T + t + t (As speed 2 and 3 have the same time periods)

We know that T = 4t, substituting this value for T,

⇒ 4t + 2t

⇒ 6t

We now have the value of the total distance and total time.

The average speed thus will be :

$\implies \sf Average\:Speed = \dfrac{24t}{6t}$

Cancelling t,

$\implies \sf Average\:speed = \dfrac{24\not t}{6\not t}$

$\implies \sf Average\:speed = \dfrac{24}{6}$

Reducing to the lowest terms,

$\implies \sf Average\:speed = 4$

Therefore the average speed = 4m/s

Answer 2

The correct answer to your question is

4m/second^2

Hope this helps you ❤️☑️