Question

In a car race, carA takes time t less than car B to cross the finishing line, but has a velocity v more than that of while crossing the finishing line. if a1 and a2 are acceleration of c ar A & B then find the relation between v, t, a1 and a2. Assume that the cars started from rest.
( kindly expain )

Answer 1

Answer on Question #79528, Physics / Other

In a car race, car A takes time $t$ less than car B to cross the finishing line, but has a velocity $v$ more than that of while crossing the finishing line. If a1 and a2 are acceleration of car A & B then find the relation between $v, t, a1$ and a2. Assume that the cars started from rest.

Solution:

For car A:

acceleration = $a_1$

total time $t_A = t_B - t$

final velocity $v_A = v_B + v$

For car B:

acceleration = $a_2$

total time = $t_B$

final velocity = $v_B$

The kinematic equation for distance:

$$s = \frac{1}{2} a_1 (t_B - t)^2 = \frac{1}{2} a_2 (t_B)^2$$

therefore

$$\frac{t_B - t}{t_B} = \sqrt{\frac{a_2}{a_1}}$$ $$1 - \frac{t}{t_B} = \sqrt{\frac{a_2}{a_1}}$$ $$t_B = \frac{t}{1 - \sqrt{\frac{a_2}{a_1}}}$$

Another kinematic equation:

$$s = \frac{v_A^2}{2 a_1} = \frac{(v_B + v)^2}{2 a_1}$$ $$s = \frac{v_B^2}{2 a_2}$$

therefore

$$\frac{v_B + v}{v_B} = \sqrt{\frac{a_1}{a_2}}$$ $$1 + \frac{v}{v_B} = \sqrt{\frac{a_1}{a_2}}$$ $$v = v _ {B} \left(\sqrt {\frac {a _ {1}}{a _ {2}}} - 1\right)$$

The kinematic equation for velocity:

$$v _ {B} = a _ {2} t _ {B}$$

So,

$$v = a _ {2} t _ {B} \left(\sqrt {\frac {a _ {1}}{a _ {2}}} - 1\right) = a _ {2} \frac {t}{1 - \sqrt {\frac {a _ {2}}{a _ {1}}}} \left(\sqrt {\frac {a _ {1}}{a _ {2}}} - 1\right) = a _ {2} t \frac {\left(\frac {\sqrt {a _ {1}} - \sqrt {a _ {2}}}{\sqrt {a _ {2}}}\right)}{\left(\frac {\sqrt {a _ {1}} - \sqrt {a _ {2}}}{\sqrt {a _ {1}}}\right)}$$ $$v = t \frac {a _ {2} \sqrt {a _ {1}}}{\sqrt {a _ {2}}} = t \sqrt {a _ {1} a _ {2}}$$

Answer: $v = t\sqrt{a_1a_2}$ .

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