Question

A car travels from A to B with
speed of 30km Ihr
return back to with a speed of 50 km /hrs find ÷
Displacement of car , Average speed of the car ,
Distance
travelled by
the car​

Answer 1

CORRECT QUESTION :-

A car travels from A to B with  speed of 30 km/h  and returns back to its initial position with a speed of 50 km/h find:

Displacement of car

Average speed of the car

Distance  travelled by  the car​

GIVEN :-

Initial velocity(u) = 30 km/h

Final velocity(v) = 50 km/h

TO FIND :-

• Displacement of car

• Average speed of the car

• Distance  travelled by  the car​

ACKNOWLEDGEMENT :-

Distance is the total length of the path between the initial point and the final position.

Displacement is the shortest length of the path between the initial point and the final position.

SOLUTION :-

Let the distance between A and B be x km.

$\bigstar$ Displacement = 0 as the body returns to its initial position.

$\rule{150}{2}$

$\bigstar$ Distance = x + x = 2x km

$\rule{150}{2}$

$\bigstar$ Time taken to travel from A to B, t₁ = Distance from A to B/Speed

$\longrightarrow$ t₁ = x/30

Time taken to travel from B to A, t₂ = Distance from B to A/Speed

$\longrightarrow$ t₂ = x/50

As we know,

$\boxed{\sf{\displaystyle{\dag\ Average\ speed=\frac{Distance\ travelled}{Total\ time\ taken} }}}$

Total time taken = t₁ + t₂

$\displaystyle{\sf{=\frac{x}{30}+\frac{x}{50} }}\\\\\\\displaystyle{\sf{=\frac{5x+3x}{150} }}\\\\\\\displaystyle{\sf{=\frac{8x}{150} }}$

Now,

$\sf{\displaystyle{ Average\ speed=\frac{2x}{\frac{8x}{150}} }}\\\\\\\sf{\displaystyle{ \implies Average\ speed=2x \times\frac{150}{8x} }}\\\\\\\sf{\displaystyle{ \implies Average\ speed=\frac{150}{4} }}\\\\\\\boxed{\boxed{\sf{\displaystyle{ \implies Average\ speed=37.5\ m/s }}}}$

$\rule{150}{2}$

Answer 2

Given:

• A car travels from A to B with  speed of 30 km/h
• Return back to A with a speed of 50 km/h

To Find:

• Displacement of Car
• Average speed of car
• Distance travelled by Car

Solution:

We know that,

• Distance is the length of actual path covered by body

• Displacement is the distance between starting and final position

➺ $\pink{\boxed{\bf{Speed=\dfrac{Distance}{Time}}}}$

➺ $\pink{\boxed{\bf{Time=\dfrac{Distance}{Speed}}}}$

➺ $\pink{\boxed{\rm{Average\:Speed=\dfrac{Total\:distance\:covered}{Total\:time\:taken}}}}$

$\rule{190}{1}$

Now,

Since, car has returned to its starting position

So,

Distance between starting and final position of car is 0km

∴ Displacement of Car= 0 km

$\rule{190}{1}$

Let the time taken by car to travel from A to B be t₁ and time taken by car to travel from B to A be t₂

Also, let the distance between A and B be x.

So,

$\longrightarrow\rm{t_{1}=\dfrac{x}{30}\:h}$

$\longrightarrow\rm{t_{2}=\dfrac{x}{50}\:h}$

Also,

Total distance covered by car= D

D= x+x= 2x km

$\rule{190}{1}$

Now,

$\longrightarrow\sf{Average\:Speed=\dfrac{Total\:distance\:covered}{Total\:time\:taken}}$

$\longrightarrow\rm{Average\:Speed=\dfrac{D}{t_{1}+t_{2}}}$

$\longrightarrow\rm{Average\:Speed=\dfrac{2x}{\dfrac{x}{30}+\dfrac{x}{50}}}$

$\longrightarrow\rm{Average\:Speed=\dfrac{2\cancel{x}}{\cancel{x}\bigg(\dfrac{1}{30}+\dfrac{1}{50}\bigg)}}$

$\longrightarrow\rm{Average\:Speed=\dfrac{2}{\dfrac{1}{30}+\dfrac{1}{50}}}$

$\longrightarrow\rm{Average\:Speed=\dfrac{2}{\dfrac{5+3}{150}}}$

$\longrightarrow\rm{Average\:Speed=\dfrac{2\times150}{8}}$

$\longrightarrow\rm{Average\:Speed=\dfrac{\cancel{300}}{\cancel{8}}}$

$\longrightarrow\rm\green{Average\:Speed=37.5\:Km/h}$