Question

A particle starts with velocity 5m/s and is retarded with 2m/s^2. Find distance travelled in 3rd second

Answer 1

Hey Dear,

◆ Answer -

Distance travelled in 3rd second = 0.25 m.

● Explanation -

Time taken for the particle to stop is calculated as -

v = u + at

0 = 5 + (-2)t

t = 5/2

t = 2.5

Distance travelled in 2 s -

s = ut + 1/2 at^2

s = 5 × 2 + 1/2 × (-2) × 2^2

s = 10 - 4

s = 6 m

Total distance travelled in 2.5 s -

s = ut + 1/2 at^2

s' = 5 × 2.5 + 1/2 × (-2) × 2.5^2

s' = 12.5 - 6.25

s' = 6.25 m

Thus, distance travelled in 3rd second (2 s -> to stopping) -

Distance = s' - s

Distance = 6.25 - 6

Distance = 0.25 m

Therefore, distance travelled in 3rd second is 0.25 m.

* Note here that particle will not move after it reaches to zero velocity.

Thanks dear...

Answer 2

Explanation:

A particle starts with velocity 5m/s and is retarded with 2m/s^2. The distance travelled in 3rd second is 0.25 m