Question

A particle moving with speed of 10m/s decelerate uniformly and comes to rest after travelling a distance of 20m. If the particle is moving with the speed of 20m/s and decelerate at the same rate what would be the stopping distance

Answer 1

${\huge {\mathfrak {Answer :-}}}$

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v = final velocity

u = initial velocity

s = Distance

a = Acceleration

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✨ First case,

v = 0 m/s

u = 10 m/s

s = 20 m.

➡️ v^2 = u^2 + 2as

Or, 0 = 100 + 40a

Or, a = - 100 /40 = - 5/2.

.°. a = - 5/2 m/s^2

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✨Second case,

v = 0 m/s

u = 20 m/s

a = - 5/2 m/s^2

➡️ v^2 = u^2 + 2as

Or, 0 = 400 - 5s

Or, s = 400/5 = 80.

.°. The stopping distance will be 80 m.

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Thanks..

Answer 2

Answer:

Explanation:

Known Terms :-

u = Initial Speed

t =  Time taken

v = Final Speed

a = Acceleration

s = Distance

Given :-

Initial speed of a particle,  u = 10 m/s  

Final   speed of a particle, v = 0 m/s  (Because it comes rest)

Distance traveled by the particles = 20 m.

To calculate :-

Stopping distance = ???

Formula to be used :-

v² = u² + 2as  

Solution :-

1st we will find the acceleration.

v² = u² + 2as  

(0)² = (10)² + 2 × a × 20

0 = 100 + 40a

a = - 100 /40

a = - 5/2.

Secondly we will put the value of acceleration on the same formula.

In this we take

v = 0 m/s

u = 20 m/s  

a = - 5/2 m/s^2

Putting all the values

v² = u² + 2as  

0 = 400 - 5s

s = 400/5

s = 80 meters.

Hence, the particle's stopping distance is 80 meters.