Question

Qq.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

Step-by-step explanation:

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)

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

For case II:-

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

t = t₁ + t₂

From equation i),

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

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

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

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

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

For case III:-

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

t = t₁ + t₂ + t₃

Again from equation i)

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

Now,

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

Answer 2

Step-by-step explanation:

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