Question

a body starts from rest and is phone to comment 32 during the fifth second of its motion find the acceleration if it is uniformly accelerated​

Answer 1

Answer:

A body starts from rest and is found to cover 3m during the 5th second of its motion. What is the acceleration if it is uniformly accelerated?

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Let’s review the 4 basic kinematic equations of motion for constant acceleration (this is a lesson – suggest you commit these to memory):

s = ut + ½ at^2 …. (1)

v^2 = u^2 + 2as …. (2)

v = u + at …. (3)

s = (u + v)t/2 …. (4)

where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

In this case, we know u = 0 and we want to find a, such that the distance covered in the 5th second is 3m, which is s(5) – s(4)

s(5) = 0 + ½a(5^2) = 12.5a

s(4) = 0 + ½a(4^2) = 8a

and we are given that s(5) – s(4) = 3

so 12.5a – 8a = 3 then a = 3/4.5 = 0.667m/s^2

The acceleration is 0.667m/s^2.

Explanation:

Answer 2

Explanation:

Let’s review the 4 basic kinematic equations of motion for constant acceleration (this is a lesson – suggest you commit these to memory):

s = ut + ½ at^2 …. (1)

v^2 = u^2 + 2as …. (2)

v = u + at …. (3)

s = (u + v)t/2 …. (4)

where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

In this case, we know u = 0 and we want to find a, such that the distance covered in the 5th second is 3m, which is s(5) – s(4)

s(5) = 0 + ½a(5^2) = 12.5a

s(4) = 0 + ½a(4^2) = 8a

and we are given that s(5) – s(4) = 3

so 12.5a – 8a = 3 then a = 3/4.5 = 0.667m/s^2

The acceleration is 0.667m/s^2.