Question

A body covers 10m in 4th second and 15m in 6th second of its uniformly accelerated motion. How far will it travel in the next 3rd seconds.​

Answer 1

Answer:

60 m

Explanation:

In nth sec, displacement = u + a(2n - 1)/2

For n = 4(4th sec), S = 10m

=> 10 = u + a(2*4 - 1)/2

=> 20 = 7a + 2u ...(1)

For n = 6(6th sec), S = 15m

=> 15 = u + a(2*6 - 1)/2

=> 30 = 11a + 2u ...(2)

Subtract (1) from (2), we get

4a = 10 → a = 2.5 m/s².

Thus, 20 = 7(2.5) + 2u → 1.25 m/s = u

Therefore, displacement in next 3 sec

= displacement in 7th + 8th + 9th sec

= [u + a(2*7 - 1)/2] + [u + a(2*8 - 1)/2] + [u + a(2*9 - 1)/2]

= u + 6.5a + u + 7.5a + u + 8.5a

= 3u + 22.5a

= 3(1.25) + 22.5(2.5)

= 60 m

Answer 2

Explanation:

A body covers 10m in 4th second and 15m in 6th second of its uniformly accelerated motion. How far will it travel in the next 3rd seconds.

Solution

In nth sec, displacement = u + a(2n - 1)/2

For n

= 4(4th sec), S = 10m

★ 10

= u + a(2*4 - 1)/2

★ 20

= 7a + 2u ...(1)

For n = 6(6th sec), S = 15m

★ 15

= u + a(2*6 - 1)/2

★ 30

= 11a + 2u ...(2)

Subtract (1) from (2), we get

4a

= 10 → a

= 2.5 m/s².

Thus, 20

= 7(2.5) + 2u

→ 1.25 m/s = u

Therefore, displacement in next 3 sec

❈ displacement in 7th + 8th + 9th sec

✻ [u + a(2*7 - 1)/2] + [u + a(2*8 - 1)/2] + [u + a(2*9 - 1)/2]

✻ u + 6.5a + u + 7.5a + u + 8.5a

✻ 3u + 22.5a

✻ 3(1.25) + 22.5(2.5)

✭ 60 m