Question

In a car race on straight road, car A takes a time ‘t’ less than car B at the finish and passes finishing point with a speed ‘v’ more than that of car B. Both the cars start from rest and travel with constant acceleration \({{\rm{a}}_1}\) and \({{\rm{a}}_2}\) respectively. Then ‘v’ is equal to:
1. \(\frac{{2{a_1}{a_2}}}{{{a_1} + {a_2}}}t\)
2. \(\sqrt {2{a_1}{a_2}} t\)
3. \(\sqrt {{a_1}{a_2}} t\)
4. \(\frac{{{a_1} + {a_2}}}{2}t\)

Answer 1

Correct Answer - Option 3 : \(\sqrt {{a_1}{a_2}} t\)

Concept:

The relation between velocity and time is a simple one during uniformly accelerated, straight-line motion. The longer the acceleration, the greater the change in velocity. Change in velocity is directly proportional to time when acceleration is constant. If velocity increases by a certain amount in a certain time, it should increase by twice that amount in twice the time. 

Calculation:

From question, we need to find the value of ‘v’ which is difference between the velocity of the car ‘A’ and ‘B’, which is:

v = v_A - v_B

The car ‘A’ takes t_o time which is less than car ‘B’ which takes ‘t + t_o’

The acceleration of car ‘A’ is given as:

\({a_1} = \frac{{{v_A}}}{{{t_o}}} \Rightarrow {v_A} = {a_1}{t_o}\)

The acceleration of car ‘B’ is given as:

\({a_2} = \frac{{{v_B}}}{{t + {t_o}}} \Rightarrow {v_B} = {a_2}\left( {t + {t_o}} \right)\)

Now, the velocity ‘v’ is:

⇒ v = a₁ t_o - a₂ (t + t_o)

⇒ v = (a₁ - a₂)t_o - a₂t

Now, the position of the car is given by the formula:

\(s = ut + \frac{1}{2}at\)

Where,

u = Initial velocity

t = Time

a = Acceleration

The position of car ‘A’ is:

\(\Rightarrow {s_A} = \left( {0 \times {t_o}} \right) + \frac{1}{2}\left( {{a_1}.t_o^2} \right)\)

\(\therefore {s_A} = \frac{1}{2}\left( {{a_1}.t_o^2} \right)\)

The position of car ‘B’ is:

\(\Rightarrow {s_B} = \left( {0 \times \left( {t + {t_o}} \right)} \right) + \frac{1}{2}\left( {{a_2}.{{\left( {{t_o} + t} \right)}^2}} \right)\)

\(\therefore {s_B} = \frac{1}{2}\left( {{a_2}.{{\left( {{t_o} + t} \right)}^2}} \right)\)

From question,

s_A = s_B

\(\Rightarrow \frac{1}{2}\left( {{a_1}.t_o^2} \right) = \frac{1}{2}\left( {{a_2}.{{\left( {{t_o} + t} \right)}^2}} \right)\)

\(\Rightarrow {a_1}.t_o^2 = {a_2}.{\left( {{t_o} + t} \right)^2}\)

Taking square root on both sides,

\(\Rightarrow \sqrt {{a_1}} .{t_o} = \sqrt {{a_2}} .\left( {{t_o} + t} \right)\)

\(\Rightarrow \sqrt {{a_1}} .{t_o} = \sqrt {{a_2}} .{t_o} + \sqrt {{a_2}} .t\)

\(\Rightarrow \sqrt {{a_1}} .{t_o} - \sqrt {{a_2}} .{t_o} = \sqrt {{a_2}} .t\)

\(\Rightarrow {t_o}\left( {\sqrt {{a_1}} - \sqrt {{a_2}} } \right) = \sqrt {{a_2}} .t\)

\(\therefore {t_o} = \frac{{\sqrt {{a_2}} .t}}{{\sqrt {{a_1}} - \sqrt {{a_2}} }}\)

Now, the velocity is:

\(\Rightarrow v = \left( {{a_1} - {a_2}} \right)\left( {\frac{{\sqrt {{a_2}} .t}}{{\sqrt {{a_1}} - \sqrt {{a_2}} }}} \right) - {a_2}t\)

\(\Rightarrow v = \left( {\sqrt {{a_1}} + \sqrt {{a_2}} } \right)\left( {\sqrt {{a_1}} - \sqrt {{a_2}} } \right)\left( {\frac{{\sqrt {{a_2}} .t}}{{\sqrt {{a_1}} - \sqrt {{a_2}} }}} \right) - {a_2}t\)

\(\Rightarrow v = \left( {\sqrt {{a_1}} + \sqrt {{a_2}} } \right)\left( {\sqrt {{a_2}} .t} \right) - {a_2}t\)

\(\Rightarrow v = \sqrt {{a_1}{a_2}} .t + {a_2}t - {a_2}t\)

\(\therefore v = \sqrt {{a_1}{a_2}} .t\)