Question

a body starting from rest moves with uniform acceleration on a horizontal surface. the body covers 3 consecutive equal distances from beginning in time t1,t2, and t3 second. The ratio of t1:t2:t3 is?

Answer 1

Answer:- $1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2})$

Explanation:

Let the acceleration of particle be a , initial velocity be u and total distance be x.

We shall use 2nd equation of motion,

S = ut + 1/2at²

Since in this case, body starts from rest, we shall take,

u = 0

Then,

S = 1/2at²

For Case I:-

S = x/3

t = t₁

So we have from equation i)

$\frac{x}{3}=\frac{1}{2}a(t_1)^2\\\;\\({t_1})^2=\frac{2x}{3a}\\\;\\t_1=\sqrt{\frac{2x}{3a}}$

For case II:-

S = x/3 + x/3 =2x/3

t = t₁ + t₂

From equation i),

$\frac{2x}{3}=\frac{1}{2}a(t_1+t_2)^2\\\;\\\frac{4x}{3a}=(t_1+t_2)^2\\\;\\(t_1+t_2)^2=\frac{4x}{3a}$

$t_1+t_2=\sqrt{\frac{4x}{3a}}$

$t_2=\sqrt{\frac{4x}{3a}}-t_1$

$\text{Putting value of}\;\;t_1$

$t_2=\sqrt{\frac{4x}{3a}}-\sqrt{\frac{2x}{3a}}$

For case III:-

S = x/3 + x/3 + x/3 = x

t = t₁ + t₂ + t₃

Again from equation i)

$x=\frac{1}{2}a(t_1+t_2+t_3)^2\\\;\\\frac{2x}{a}=(t_1+t_2+t_3)^2\\\;\\t_1+t_2+t_3=\sqrt{\frac{2x}{a}}\\\;\\\text{Putting vaules of}\;t_1\;\text{and}\;t_2\\\;\\\sqrt{\frac{2x}{3a}}+\sqrt{\frac{4x}{3a}}-\sqrt{\frac{2x}{3a}}+t_3=\sqrt{\frac{2x}{a}}\\\;\\\sqrt{\frac{4x}{3a}}+t_3=\sqrt{\frac{2x}{a}}\\\;\\t_3=\sqrt{\frac{2x}{a}}-\sqrt{\frac{4x}{3a}}$

Now,

$t_1:t_2:t_3\\\;\\=\sqrt{\frac{2x}{3a}}:\sqrt{\frac{4x}{3a}}-\sqrt{\frac{2x}{3a}}:\sqrt{\frac{2x}{a}}-\sqrt{\frac{4x}{3a}}\\\;\\=\sqrt{\frac{2}{3}}:(\sqrt{\frac{4}{3}}-\sqrt{\frac{2}{3}}):(\sqrt{2}-\sqrt{\frac{4}{3}})\\\;\\\text{Multiplying by}\sqrt{3}\\\;\\=\sqrt{2}:(2-\sqrt{2}):(\sqrt{6}-2)\\\;\\\text{Dividing by}\;\sqrt{2}\\\;\\=1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2})$

Answer 2

$\huge{\underline{\underline{.........Answer.........}}}$

$\huge{\underline{Given:-}}$

${S_1 = S_2 = S_3 = S}$

$\huge{\underline{Explanation:-}}$

u = 0 m/s.

a = constant

• Case-1

From second kinematic equation.

${S = S_1 = \frac {1}{2} a({t_1})^2}$

${ \therefore t_1 = \sqrt{ \frac{2S}{a}}}$

• Case-2

From second kinematic equation.

As the particle has travelled for the same distance two times the equation becomes.

${2S = \frac{1}{2}a({t_1 + t_2})^2}$

${ \therefore t_1 + t_2 = \sqrt{2} \times \sqrt{ \frac{2s}{a} } }$

Now it becomes,

${t_2 = \sqrt{2} \times \sqrt{ \frac{2S}{a}} - \sqrt{ \frac{2S}{a}}}$

${t_2 = \sqrt{ \frac{2S}{a}} \times( \sqrt{2} - 1)}$

• Case-3

As the particle has travelled for the same distance three times the equation becomes.

${3S = \frac{1}{2}a({t_1 + t_2 + t_3})^2}$

${ \therefore t_1 + t_2 + t_3 = \sqrt{3} \times \sqrt{ \frac{2S}{a}}}$

Now it becomes,

${t_3 = (\sqrt{3} \times \sqrt{ \frac{2S}{a}}) - \sqrt{ \frac{2S}{a}} - (\sqrt{ \frac{2S}{a}} \times( \sqrt{2} - 1))}$

${t_3 = \sqrt{ \frac{2S}{a}}( \sqrt{3} - \sqrt{2})}$

Now ratio

${t_1:t_2:t_3 = \sqrt{ \frac{2S}{a}}:\sqrt{ \frac{2S}{a}}( \sqrt{2} - 1): \sqrt{ \frac{2S}{a}}( \sqrt{3} - \sqrt{2})}$

${t_1:t_2:t_3 = 1:( \sqrt{2} - 1):( \sqrt{3} - ( \sqrt{2})}$

$\huge{\boxed{\boxed{t_1:t_2:t_3 = 1:( \sqrt{2} - 1):( \sqrt{3} - ( \sqrt{2})}}}$