Question

a particle initially at rest moves with an acceleration of 8 m / sec ^2 for 8 secs find the distance travelled in (I) 7 sec (ii) 8 sec (iii) 8 th second
subject - physics
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Answer 1

Answer:

Distance traveled in 5th second = Distance traveled in 5 s− Distance traveled in 4 s

Using s=ut+(1/2)at

2

Distance traveled in 5 s =(1/2)×5×5

2

=62.5 m

Distance traveled in 4 s =(1/2)×5×4

2

=40 m

Distance traveled in 5th s =(1/2)×5×5

2

=62.5 m=S

2

−S

1

=(62.5−40)m=22.5m

Answer 2

GIVEN

• Initial Velocity = 8m/s²
• Time = 8 sec

To FiND

• Distance at 7 sec , 8 sec

SOLUTION

We know that

S = ut + 1/2 at²

where

• S = displacement
• u = initial Velocity
• a = acceleration
• t = time

S at 7 sec = 8*7 + 1/2 × 8 × 7²

= 56 + 196

= 252 m

S at 8sec = 8*8 + 1/2 × 8 × 8²

= 64 + 256

= 320 m