Question

A particle travels 10 m in first 5 seconds and 10 m in next 3 seconds. Assuming constant acceleration, what is the distance travelled by the particle in next 3 seconds? (AMU +2 Entrance Test. Science)

$$$BRAINLY BENEFACTOR$$$

Answer 1

We know that initial velocity of a particle will be 0.

(1)

Given time = 5 seconds, Distance = 10m/s.

S = ut + 1/2 a * t^2.

10 = u * 5 + 1/2 * a * (5)^2

10 = 5u + 25a/2

10 - (5u + 25a/2) = 0

10 - (10u + 25a) = 0

10 - 10u + 25a = 0

-5(2u + 5a - 4) = 0

2u + 5a - 4 = 0

2u + 5a = 4   ---------------- (1)



(2)

Given t = 5 + 3 = 8 seconds, distance = 20m.

S = ut + 1/2 a * t^2

20 = 8u + 1/2 * a(8)^2

20 = 8u + 64a/2

20 - (8u + 64a/2) = 0

20 - 8u - 32a = 0

-4(2u + 8a - 5) = 0

2u + 8a - 5 = 0

2u + 8a = 5               -------------------- (2)


On solving (1) and (2), we get

4 - 5 = 2u + 5a - 2u - 8a

-1 = -3a

a = 1/3        ------------------------- (3)


Substitute (3) in (2), we get

2u + 8a = 5

2u + 8(1/3) = 5

2u + 8/3 = 5

6u + 8 = 15

6u = 7

u = 7/6    ------------------- (4).


Therefore the distance covered in first 10 seconds

S = ut + 1/2 at^2

   = 7/6 * 10 + 1/2 * 1/3(10)^2

   = 70/6 + 1/6 * 100

   = 70/6 + 100/6

   = 170/6

   = 28.3m.


Therefore the distance traveled by the particle in next 3 seconds 

 = 28.3 - 20

= 8.3m.


Hope this helps!

Answer 2

Answer:

Explanation:

Hope it will help u .