Physics, asked by madhupenta77, 23 days ago

A particle starts form rest accelerates at 2m/s2
for 10sec and then goes for constant speed
for 30sec and then decelerates 4m/s2
till its stops. The distance traveled by the particle is

Answers

Answered by tirthnagar3
1

Answer:

Displacement in first, 10sec

s

1

=

2

1

a

1

t

1

2

=

2

1

×2×10

2

=100m

Speed achieve in first 10sec

v

1

=u

1

+a

1

t

1

=0+2×10=20m/s

Displacement in 30sec, acceleration is zero.

s

2

=v

1

t

2

=20×30=600m

Displacement in last due to deacceleration, a

3

=−4m/s

2

v

2

−v

1

2

=2as

s

3

=

2×(−4)

0−20

2

=50m

Net displacement

S

total

=s

1

+s

2

+s

3

=100+600+50=750m

Total displacement is 750m

Explanation:

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