Question

A particle starts from rest accelerated 2 metre per second square for 10 seconds and then goes for contacts feed for 30 seconds and then disocal rates at 4 metre per second square till it stops what is the distance travelled by it

Answer 1

$\Huge{\underline{\underline{\mathfrak{Answer \colon }}}}$

Case I

The particle moves with an acceleration 2m/s²

From the Question,

• Initial Velocity,u = 0 m/s

• Time Taken,t = 10s

To find

Distance Travelled

Using the Relation,

$\huge{ \boxed{ \boxed{ \mathtt{s = ut + \frac{1}{2} at {}^{2} }}}}$

Putting the values,we get :

$\sf{s = 0(10) + \frac{1}{2}.2.(10) {}^{2} } \\ \\ \rightarrow \: \sf{s = 10 {}^{2} } \\ \\ \rightarrow \: \boxed{ \sf{ \green{s = 100 \: m}}}$

Final Velocity at the end of this interval is equal to the initial velocity of the particle at the starting of next interval

Implies,

$\sf{v = u + at} \\ \\ \implies \: \sf{v = 0 + 2(10)} \\ \\ \huge{\implies \: \sf{v = 20 ms {}^{ - 1} }}$

Case 2

When the particle is moving with an acceleration - 4 m/s²

From the Question,

• Initial Velocity,u = 20 m/s

• Time,t = 30s

• Final Velocity,v = 0 m/s

Putting the values,we get :

$\sf{{v}^{2} - {u}^{2} = 2as} \\ \\ \implies \: \sf{- 8s = - {20}^{2}} \\ \\ \implies \: \boxed{ \green{\sf{s = 50 \: m}}}$

The distance covered during the whole journey would be 150 m

Answer 2

$\huge{\text{\underline{Correct\:Question:-}}}$

A particle starts from rest accelerates at 2m/s² for 10s and then goes for constant speed for 30s and then accelerates at 4m/s² till it stops. What is the distance travelled by it?

$\huge{\text{\underline{Solution:-}}}$

★ Given:-

• t = 10s
• a = 2m/s²
• u = 0
• v = 20m/s

★ For first 10 sec

$\implies$ S1 = 1/2 × a × t²

$\implies$ S1 = 1/2 × 2 × 10 × 10

$\implies\boxed{S1 = 100m}$

★ Velocity reached:-

$\implies$v = u + at

$\implies$v = 0 + 2 × 10

$\implies\boxed{v = 20m/s}$

★ For next 30s :-

$\implies$S2 = 20 × 30

$\implies\boxed{S2 = 600m}$

★ Under retardation of 4m/s²:-

$\implies$v² = 2a × S3

$\implies$ (20)² = 2 × 4 × S3

$\implies$ 400 = 8 × S3

$\implies$ S3 = 400/8

$\implies\boxed{S3 = 50m}$

★ Total distance covered:-

$\implies$S1 + S2 + S3

$\implies$100 + 600 + 50

$\implies\boxed{= 750m}$

Therefore,the distance covered by the particle is 750m.

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