Question

a body start from rest and moving with uniform acceleration of 4m/s2 cover half of its total path during the last second of its motion. Find the total time taken and the total distance covered​

Answer 1

Answer:

Total distance moved by it is 23.2 m and total times taken by it is 3.41 s

Explanation:

Let the total time during the motion is given as

$t = N s$

now the distance moved by it in this time

$d = 0 + \frac{1}{2}(4) N^2$

$d = 2N^2$

now half of this total distance is moved in last second of the motion

so we have

$N^2 = \frac{1}{2}(4)N^2 - \frac{1}{2}(4)(N - 1)^2$

$N^2 = 2(N^2 - (N - 1)^2)$

$N^2 = 2(2N - 1)$

$N^2 - 4N + 2 = 0$

$N = 3.41 s$

And total distance is given as

$d = 2(3.41)^2$

$d = 23.2 m$

#Learn

Topic : Kinematics

https://brainly.in/question/126834