Question

In a car race, car $A$ takes $20\text{s}$ less than car $B$ to finish. It passes the finishing point with speed $v$ more than that of car $B$. Assuming that both the cars start from rest and travel with a constant acceleration of $50{\text{m/s}}^2$ and $40{\text{m/s}}^2$ respectively, what is the value of $v$ ?

• A. $620\text{m/s}$
• B. $894\text{m/s}$
• C. $682\text{m/s}$
• D. $864\text{m/s}$

Answer 1

The correct option is B $894\text{m/s}$

Given that,

Acceleration of car $A,a_1=50{\text{m/s}}^2$

Acceleration of car $B,a_2=40{\text{m/s}}^2$

Initial velocity of both cars, $u=0\text{m/s}$

Let car $A$ take $t_1$ time to travel from rest to destination, reaching with velocity $v_1$.

We know that

$v=u+at$

$\Rightarrow v_1=50{\text{m/s}}^2\times t_1\dots \dots \left(i\right)$

Time taken by car $B$,

$t_2=t_1+20\text{sec}$

Velocity of car $B$, $v_2=v_1-v$

Velocity of car $B$ is given by

$v_2=a_2{t}_2$ (from equation $\left(i\right)$)

$\Rightarrow v_1-v=a_2(t_1+20)$

$\Rightarrow v_1-v=40(t_1+20)\dots \dots \left(ii\right)$

Subtracting equation $\left(ii\right)$ from equation $\left(i\right)$

$\Rightarrow v_1-(v_1-v)=50t_1-40(t_1+20)$

$\Rightarrow v=10t_1-800\dots \dots \left(iii\right)$

Now, total distance travelled by both the cars is the same.

So, $S_1=S_2$

From $S=ut+\frac{1}{2}a{t}^2$

$\frac{1}{2}\times 50\times t_1^2=\frac{1}{2}\times 40\times (t_1+20{)}^2$

$\Rightarrow 25\times t_1^2=20\times (t_1+20{)}^2$

$\Rightarrow 25t_1^2=20\times (t_1^2+40t_1+400)$

$\Rightarrow (25-20)t_1^2-800t_1-8000=0$

Solving the quadratic equation, we get:

$t_1=169.44$ and $-9.44$ (Neglected)

Substitute $t_1$ in equation $\left(iii\right)$

$\Rightarrow v=10t_1-800$

$\Rightarrow v=10\times 169.44-800$

$\Rightarrow v=894.4\text{m/s}\cong 894\text{m/s}$