Question

a particle starting from rest and moving with unifrom acceleration of 0.3m/s2 descri (9/25)th of the whole distance in the last second of its motion. How long has it been in motion and what is the whole distance?​

Answer 1

Correct Question:

• A particle starting from rest and moving with uniform acceleration of 0.3 m/s² describes 9/25 th of the whole distance in the last second of its motion. How long has it been in motion and what is the whole distance ?

Answer:

• Time taken = 5 seconds
• Distance covered = 3.75 m

Given Information:

• Initial Velocity (u) = 0
• Acceleration (a) = 0.3 m/s²
• Distance covered in the last second = (9/25)th part of complete distance.

To find:

• Time taken to complete the whole distance
• Distance covered by the particle

Steps:

We know that, According to second equation of motion,

$\boxed{ s = ut + \dfrac{1}{2}at^2}$

where, 's' refers to the path length, 'u' refers to initial velocity, 't' refers to the time and 'a' refers to the acceleration.

Calculating the distance covered in between 't' and 't-1' seconds we get:

$\implies s_s_{(t-1)} = s _{(t)} - u(t-1) - \dfrac{1}{2}a(t-1)^2\\\\\\\implies \dfrac{9}{25}(s_{(t)}) = s_{(t)} - \dfrac{1}{2}a(t^2 - 2t + 1 )\\\\\\\implies s_{(t)} - \dfrac{9}{25}(s_{(t)}) = \dfrac{1}{2} a(t^2 - 2t + 1 )\\\\\\\implies \dfrac{25-9}{25} s_{(t)} = \dfrac{1}{2} a(t^2 - 2t + 1)\\\\\\\implies \dfrac{16}{25} s_{(t)} = \dfrac{1}{2}a(t^2 - 2t + 1)\\\\\\\text{Substituting the value of } \:s_{(t)}\: \text{ we get,}$

$\implies \dfrac{16}{25} \times \dfrac{1}{2} at^2 = \dfrac{1}{2}a(t-1)^2\\\\\\\implies \dfrac{16}{25} t^2 = (t-1)^2\\\\\\\implies (\dfrac{4t}{5})^2 = (t-1)^2\\\\\\\text{Taking Square root on both sides we get:}\\\\\\\implies \dfrac{4t}{5} = t - 1\\\\\\\implies 1 = t - \dfrac{4t}{5}\\\\\\\implies 1 = \dfrac{t}{5}\\\\\\\implies \boxed{t = 5 \:s }$

Hence the time for which the particle was in motion is 5 seconds.

Distance covered in t = 5 seconds is calculated as:

$\implies s_{(t=5)} = ut + \dfrac{1}{2} at^2\\\\\\\text{Substituting the values, we get:}\\\\\\\implies s_{(t = 5)} = 0 ( 5 ) + \dfrac{1}{2} \times 0.3 \times (5)^2\\\\\\\implies s_{(t=5)} = 0.5 \times 0.3 \times 25\\\\\implies \boxed{\textbf{Distance covered} = 3.75 \: m}$