Question

Please answer fast because my exam will at 11:00 am !!

What will the average speed .

Give step by step but fast!!


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Answer 1

CORRECT QUESTION :-

➠ A boy goes from his school to school by bus at a speed of 20 km/h and returns back through the same route at a speed of 30 km/h . the average speed of his journey is.

TO FIND :-

➠ The average speed.

SOLUTION :-

Case 1 :-

Speed = 20 km/h

➣ Let the Distance be x km.

➣ Now apply the formula of time taken,

✮ Time = Distance/speed

➣( x km/20 km) h

➣ Time = x/20 h

Hence the Time in case 1 is x/20 hours.

Case 2 :-

Speed = 30 km/h

➣ Let the Distance be x km.

➣ Now apply the same formula of time taken,

✮ Time = Distance/speed

➣ (x km/30 km) ×h

➣ Time = x/30 h

Hence the time in case 2 is x/30 hours.

➠ As we know that the formula of average speed so let's apply,

✮ Average speed = Total Distance Traveled/Total Time

➣ [( x + x)/(x/20 + x/30)]

➣ [2x/(3x + 2x)/60]

➣ [2x/(x/12)]

➣ 2x × 12/x

➣ 12 × 2

➣ 24 km/h

Hence the average speed is 24 km/hour

Answer 2

☯ To Find:

☞The average speed of the boy.

☯ Given:

• Speed $\mathtt{\rightarrow v_{1} = 20km\:h^{-1}}$

• Speed $\mathtt{\rightarrow v_{2} = 30km\:h^{-1}}$

☯ We Know:

➝ speed :

$\boxed{speed = \dfrac{Distance}{time}}$

➝ Average speed :

$\boxed{Average\:speed = \dfrac{s_{1} + s_{2}}{t_{1} + t_{2}}}$

☯ Concept:

ATQ the distance is same in both the cases so ,we can denote the distance by a single variable .

Let the variable be x in both the cases.

so by the the given speed and taken distance ,we can find the time in both the cases.

So , we will get the average speed as:

$\boxed{\Rightarrow Average\:speed = \dfrac{x + x}{t_{1} + t_{2}}}$

$\boxed{\Rightarrow Average\:speed = \dfrac{2x}{t_{1} + t_{2}}}$

☯ Solution:

✧ To Find the time :

• For , speed $s_{1}$ ,

• Speed = $20 kmh^{-1}$
• distance = x m

➝ From the formula:

$\sf{speed = \dfrac{Distance}{time}}$

➝ We Get,

$\sf{time \rightarrow t_{1} = \dfrac{Distance}{speed}}$

putting the value,

$\sf{time \rightarrow t_{1} = \dfrac{x}{20} h}$

• For , speed $s_{2}$ ,

Speed = $30 kmh^{-1}$

distance = x m

➝ From the formula:

$\sf{speed = \dfrac{Distance}{time}}$

➝ We Get,

$\sf{time \rightarrow t_{2} = \dfrac{Distance}{speed}}$

putting the value,

$\sf{time \rightarrow t_{2} = \dfrac{x}{30} h}$

☞ Finding The average speed :.

➝ We know ,

• $s_{1} = s_{2} = x$
• $t_{1} = \dfrac{x}{20} h$
• $t_{2} = \dfrac{x}{30} h$

➝ We also know ,

$\boxed{Average\:speed = \dfrac{2x}{t_{1} + t_{2}}}$

➝ Putting the value in the formula , we get:

$\Rightarrow Average\:speed = \dfrac{2x}{\dfrac{x}{20} + \dfrac{x}{30}}$

$\Rightarrow Average\:speed = \dfrac{2x}{\dfrac{3x + 2x}{60}}$

$\Rightarrow Average\:speed = \dfrac{2x}{\dfrac{5x}{60}}$

$\Rightarrow Average\:speed = \dfrac{2x}{5x}\times 60$

$\Rightarrow Average\:speed = \dfrac{2x}{\cancel{5x}} \times \cancel{60}$

$\Rightarrow Average\:speed = \dfrac{\cancel{2x}}{\cancel{x}} \times 12$

$\Rightarrow Average\:speed = 2 \times 12$

$\Rightarrow Average\:speed = 24 kmh^{-1}$

$\boxed{\therefore Average\:speed = 24 kmh^{-1}}$

Hence , the average speed of the boy is $24kmh^{-1}$

Extra Information:

• $v = u + at$

• $s = u + \dfrac{1}{2}at^{2}$

Where,

• v = final velocity
• u = initial velocity
• t = Time taken
• a = acceleration