Question

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Answer 1

Question:-

☛A particle starts from rest and has an acceleration of 2m/s²for 10sec. After that, it travels for 30sec with constant speed and then undergoes a retardation of 4m/s² and comes back to rest. The total distance covered by the particle is

Given:-

☛Initial speed is 0m/s

☛Has acceleration is 2m/s²

☛Time is 10 sec

Then,

☛Time travels for 30 sec

☛Constant speed

then,

☛Final speed is 0m/s

☛retardation is 4m/s²

Find:-

☛Total Distance covered-?

Formula to be used:-

☛ $\sf{s= ut+\dfrac{1}{2}at^2}$  

☛ $\sf{v^2 - u^2 = 2as}$

where,

s=distance

u=initial speed

a=acceleration

t=time

v=final peed

Solution:-

☛Initially,

by using 2nd equation,

$\sf{s= ut+\dfrac{1}{2}at^2}$

$\sf{s= 0\times 10+\dfrac{1}{2}\times 2 \times 10^2}$

$\sf{s= \dfrac{1}{2}\times 2\times 100}$

$\sf{s=100m}$

here distance s = 100m

 ______________________________

☛Then,

v=20

t=30

Traveling for 30sec with constant speed,

so here distance covered is,

$\sf{s' =vt}$

$\sf{ s' = 20\times 30= 600m}$

Distance s'= 600m

  _______________________________

☛Now,

Travelling due to retardation,

By using 3rd equation,

$\sf{v^2 - u^2 = 2as}$

$\sf{(0)^2 - (20)^2 = 2 \times (- 4) \times s''}$

s'' = 50 m

 ________________________________

Now, the total distance covered,

S= s+s'+s''

S= (100 + 600 + 50) m

S=  750 m

Hence, Distance covered by the particle is 750 m.

Answer 2

Answer:

hope so it will be useful to you mam