Question

In a car race, car A takes time t less than car B and passes the finishing point with a velocity v more than the velocity with which car B passes the point. Assuming that the cars start from rest and travel with constant accelerations a1 and a2, show that =√1 2.

Answer 1

In a car race, car A takes time t less than car B.

Let us assume that car A takes time (t' - t).

Therefore, time taken by car B is (t').

Also given that, car A passes the finishing point with a velocity v more than the velocity with which car B passes the point.

So, assume that the velocity of car B is u.

Therefore, the velocity of car A is (u + v).

Now, we know that,

⇒ v² - u² = 2as

As, u = 0

So, v = √(2as)

For car A:

→ v + u = √(2a1s) ...............(1)

For car B:

→ u = √(2a2s)......................(2)

Now,

→ v + u - u = √(2a1s) - √(2a2s)

→ v = √(2a1s) - √(2a2s).......(3)

Using the Third Equation of Motion.

⇒ s = ut + 1/2 at²

As u = 0

So,

→ s = 1/2 at²

→ t = √(2s/a)

For car A:

→ (t'-t) = √(2s/a1)

For car B:

→ t' = √(2s/a2)

Now,

→ t' - t' + t = √(2s/a2) - √(2s/a1)

→ t = √(2s/a2) - √(2s/a1)..........(4)

On Dividing v by t (means equation 3 by equation 4) we get,

→ v/t = [√(a1) - (√a2) (√a1 √a2)]/[√(a1) - √(a2)]

→ v/t = (√a1 √a2)

→ v = t[√(a1 a2)]

Answer 2

Given:-

• $\text{In a car race, car A takes time `t' less than car B}\text{ and passes the finishing}\\\text{ point with a velocity `v' more than the velocity with which car B passes}\\\text{the point.}$

• $\text{The cars start from rest and travel with constant accelerations }a_1\ and\ a_2.$

To prove:-

$\mathbf{v=(\sqrt{a_1a_2})t }$

Solution:-

A/q,

car A takes time $t_1$ seconds less than car B. Then the time taken by the car B becomes $(t_1+t)$ seconds.

Now,

let $v_1$ m/s be the velocity of B at finishing point, so the velocity of A becomes $(v_1+v)$ m/s.

Using the formula:

v + u = at

initial velocity(u) is zero as the cars start from rest.

• 1st equation for car A becomes:-

$\mathbf{v_1+v=a_1t_1}$                      .....(i)

• 2nd equation for car B becomes:-

$\bold{v_1=a_2(t_1+t)}$                   ......(ii)

Now, subtracting eq. (ii) from eq. (i);

We get:

$\mathbf{v=a_1t_1-a_2t_1-a_2t}\\\mathbf{\implies v=t_1(a_1-a_2)-a_2t}$       .....(iii)

We know that, car A and car B travel the same distance($s=s_1$).

Using the formula:

s = ut + 1/2 at²

As u = 0 for both the cars, so ut = 0.

• For car A:

$\bold{s=\frac{1}{2}a_1t_1^2}$                           .....(iv)

• For car B:

$\bold{s_1=\frac{1}{2}a_2(t+t_1)^2}$               ......(v)

• Eq.(iv) = eq. (v).

$\bold{\frac{1}{2} a_1t_1^2=\frac{1}{2}a_2(t_1+t)^2}$

$\bold{\implies t_1=\frac{\sqrt{a_2}t }{\sqrt{a_1}-\sqrt{a_2} } }$

Substituting this value of $t_1$ in equation (iii), we get:

$\bold{v=\bold{ \frac{\sqrt{a_2}t }{\sqrt{a_1}-\sqrt{a_2} }(a_1-a_2)-a_2t }}\\\\\text{On solving,}\\\\\boxed{\implies v=\sqrt{a_1a_2}t }$