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Question: A point moves with uniform acceleration and \({{v}_{1}},{{v}_{2}},{{v}_{3}}\) denote the average vel...

A point moves with uniform acceleration and v1,v2,v3{{v}_{1}},{{v}_{2}},{{v}_{3}} denote the average velocities in three successive intervals of time t1,t2,t3{{t}_{1}},{{t}_{2}},{{t}_{3}}. Then, the relation v1v2v2v3\dfrac{{{v}_{1}}-{{v}_{2}}}{{{v}_{2}}-{{v}_{3}}} is

A. t1t2t2+t3\dfrac{{{t}_{1}}-{{t}_{2}}}{{{t}_{2}}+{{t}_{3}}}
B. t1+t2t2+t3\dfrac{{{t}_{1}}+{{t}_{2}}}{{{t}_{2}}+{{t}_{3}}}
C. t1t2t1+t3\dfrac{{{t}_{1}}-{{t}_{2}}}{{{t}_{1}}+{{t}_{3}}}
D. t1t2t2t3\dfrac{{{t}_{1}}-{{t}_{2}}}{{{t}_{2}}-{{t}_{3}}}

Explanation

Solution

Hint: Assume the initial velocity, and find the final velocity after each interval in time. Find the average velocity by taking the mean of the initial and final velocity at each interval. Calculate the required value from the average velocity expressions.

Formula Used:
We can write the following equation,
v=u+atv=u+at
u is the initial velocity of the particle
v is the final velocity of the particle
a is the acceleration of the particle
t is the time interval

Complete step by step answer:
Let’s assume that the initial velocity be u.

So, we can write the equation for time interval t1{{t}_{1}},
v=u+at1{{v}^{'}}=u+a{{t}_{1}}

After the second interval the velocity is,
v=u+a(t1+t2){{v}^{''}}=u+a({{t}_{1}}+{{t}_{2}})

After the third interval the velocity is,
v=u+a(t1+t2+t3){{v}^{'''}}=u+a({{t}_{1}}+{{t}_{2}}+{{t}_{3}})

As the particle is having uniform acceleration, the average velocity can be calculated by the mean of initial and final velocity in that time interval.

Hence, we can write the following three equations:
v2=u+v2=u+u+at12=u+12at1{{v}_{2}}=\dfrac{u+{{v}^{'}}}{2}=\dfrac{u+u+a{{t}_{1}}}{2}=u+\dfrac{1}{2}a{{t}_{1}}................(1)
v2=v+v2=u+at1+u+a(t1+t2)2=u+at1+12at2{{v}_{2}}=\dfrac{{{v}^{'}}+{{v}^{''}}}{2}=\dfrac{u+a{{t}_{1}}+u+a({{t}_{1}}+{{t}_{2}})}{2}=u+a{{t}_{1}}+\dfrac{1}{2}a{{t}_{2}}................(2)
v2=v+v2=u+a(t1+t2)+u+a(t1+t2+t2)2=u+at1+at2+12at3{{v}_{2}}=\dfrac{{{v}^{''}}+{{v}^{'''}}}{2}=\dfrac{u+a({{t}_{1}}+{{t}_{2}})+u+a({{t}_{1}}+{{t}_{2}}+{{t}_{2}})}{2}=u+a{{t}_{1}}+a{{t}_{2}}+\dfrac{1}{2}a{{t}_{3}}................(3)

So, subtracting equation (2) from equation (1) we get,
v1v2=12a(t1+t2){{v}_{1}}-{{v}_{2}}=-\dfrac{1}{2}a({{t}_{1}}+{{t}_{2}}).................(4)

Subtracting equation (3) from equation (2) we get,
v2v3=12a(t2+t3){{v}_{2}}-{{v}_{3}}=-\dfrac{1}{2}a({{t}_{2}}+{{t}_{3}})................(5)

Dividing equation (5) with equation (4) we get,
v1v2v2v3=t1+t2t2+t3\dfrac{{{v}_{1}}-{{v}_{2}}}{{{v}_{2}}-{{v}_{3}}}=\dfrac{{{t}_{1}}+{{t}_{2}}}{{{t}_{2}}+{{t}_{3}}}

So, the correct answer is (B).

Note: The particle is travelling with an uniform acceleration. That is why the average velocity can be found using the final and initial velocity of the time period. If the acceleration was variable, we could not use the simple mean value.
Assuming the initial velocity of the particle is not important but it is necessary to make the answer generalized. So, the initial velocity was assumed to be u.