Question

a car is traveling at 72km/h has its velocity reduced to 36km/h in 5s. If the retardation is uniform, find how much distance it has covered during this time. How much farther would it travel before coming to rest assuming the same uniform retardation?
​

Answer 1

Answer:

• Car will travel a distance of 75 metres while retarding for 5 seconds.
• And, after 5 seconds Car will travel 25 metres more to come to rest.

Explanation:

Given

• Initial velocity of Car = 72 km/h = 72 × 5/18  m/s = 20 m/s
• Final velocity of Car after retarding for a time 5 seconds = 36 km/h = 36 × 5/18  m/s = 10 m/s

To find

• Distance covered by Car in this time of 5 seconds, s₁ = ?
• How much more distance would it travel before coming to rest assuming the same uniform retardation, s₂ = ?

So

Calculating the acceleration of Car (a = ?)

By first equation of motion

→v = u + a t

[ where v represent final velocity, u represent initial velocity, a denotes acceleration and t represents time taken ]

Putting values for considered case

→ 10 = 20 + a ( 5 )

→ 10 - 20 = 5 a

→ 5 a = -10

→ a = -2 m/s²

Now,

Calculating distance covered by Car in retarding for 5 seconds (s₁ = ?)

By second equation of motion

→ s = u t + 1/2 a t²

[ Where s represent distance covered, u as initial velocity, t represent time taken and a represent acceleration ]

Putting convenient values

→ s₁ = ( 20 ) ( 5 ) + 1/2 ( -2 ) ( 5 )²

→ s₁ = 100 - 25

→ s₁ = 75 m

Calculating the more distance Car need to travel before coming to rest (s₂ = ?)

Using Third equation of motion

→ 2 a s = v² - u²

[ Where a is acceleration, s represent distance covered, v represent final velocity and u represent initial velocity ]

Putting values for considerate case

[ taking initial velocity in this case as 10 m/s ]

→ 2 ( -2 ) s₂ = ( 0 )² - ( 10 )²

→ -4 s₂ = 0 - 100

→ 4 s₂ = 100

→ s₂ = 25 m

HENCE,

• Car will travel a distance of 75 metres while retarding for 5 seconds.
• And, after 5 seconds Car will travel 25 metres more to come to rest.

Answer 2

$\sf{\blue{\underline{\underline{\pink{GIVEN:-}}}}}$

• A car is traveling at 72km/h .

• In 5s, it's velocity reduced to 36km/h .

• It's retardation is uniform .

$\sf{\blue{\underline{\underline{\pink{TO\: FIND:-}}}}}$

1. How much distance it has covered during this time[5s] .
2. How much farther would it travel before coming to rest assuming the same uniform retardation .

$\sf{\blue{\underline{\underline{\pink{SOLUTION:-}}}}}$

$\orange\star\:\bf{\gray{\underline{\pink{\boxed{\purple{v\:=\:u\:+\:at\:}}}}}}$

Where,

• v = final velocity = 36 km/h = 36 × 5/18 = 10m/s

• u = initial velocity = 72 km/h = 72 × 5/18 = 20m/s

• a = acceleration = ?

• t = time = 5s

=> 10 = 20 + a × 5

=> a × 5 = 10 - 20

=> a = -10/5

=> a = -2m/s²

[Note :- Here, -ve sign indicates that the car is retarding in motion .]

$\green\star\:\bf{\gray{\underline{\pink{\boxed{\purple{S\:=\:ut\:+\:\dfrac{1}{2}\:at^2\:}}}}}}$

Where,

• S = distance = ?

• u = initial velocity = 20m/s

• t = time = 5s

• a = acceleration = -2m/s²

$\rm{\implies\:S\:=\:20\times{5}\:+\:\dfrac{1}{2}\times{-2}\times{5^2}\:}$

$\rm{\implies\:S\:=\:100\:-\:25\:}$

$\rm\green{\implies\:S\:=\:75\:m}$

$\red{\therefore}$ It was covered “ 75m ” during 5s .

☃️ Now according to the question, it is given that

• v = final velocity = 0m/s

$\blue\star\:\bf{\gray{\underline{\pink{\boxed{\purple{v^2\:-\:u^2\:=\:2as'\:}}}}}}$

• u = 10m/s

• s' = distance = ?

• a = acceleration = -2m/s

=> 0² - (10)² = 2 × (-2) × s'

=> 0 - 100 = -4 × s'

=> -4 × s' = -100

=> 4 × s' = 100

=> s' = 100/4

=> s' = 25m

$\red{\therefore}$ It will travel “ 25m ” distance before coming to rest assuming the same uniform retardation .