Question

In a car race, car $A$ takes $20$ seconds less than car $B$ to finish and passes the finishing point with speed $v$ more than that of car $B$. Assuming that both 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 $A\left(a_1\right)=50{\text{m/s}}^2$
Acceleration of car $B\left(a_2\right)=40{\text{m/s}}^2$
Initial velocity of both cars, $\left(u\right)=0\text{m/s}$
Let car $A$ take $t_1$ time to travel from rest to destination, reaching with velocity $v_1$.
We know that
$v=at$
$\Rightarrow v_1=50{\text{m/s}}^2\times t_1\dots \dots \left(i\right)$
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 are same.
So, $S_1=S_2$
$\frac{1}{2}\times 50\times t_1^2=\frac{1}{2}\times 40\times (t_1+20{)}^2$
$\Rightarrow 25\times t_1^2=20\times (t_1+20{)}^2$
$\Rightarrow 25t_1^2=20\times (t_1^2+40t_1+400)$
$\Rightarrow (25-20)t_1^2-800t_1-8000=0$
Solving the quadratic equation, we get:
$t_1=169.44$ and $-9.44$ (Neglected)
Substitute $t_1$ in equation $\left(iii\right)$
$\Rightarrow v=10t_1-800$
$\Rightarrow v=10\times 169.44-800$
$\Rightarrow v=894.4\text{m/s}\cong 894\text{m/s}$