Question

Two cars $A$ and $B$ are moving in same direction. Car $A$ is moving with speed $10m/s$ and at a distance $80m$ behind car $B$ which starts from rest and with uniform acceleration $a_o$. The maximum value of $a_o$ for which car $A$ can reach car $B$

• A. $\frac{1}{8}m/s^2$
• B. $\frac{7}{8}m/s^2$
• C. $\frac{5}{16}m/s^2$
• D. $\frac{3}{8}m/s^2$

Answer 1

The correct option is C $\frac{5}{16}m/s^2$

Suppose after time $t$ the car A reaches the the car B then distance traveled by B i.e. $x_B$ and the distance by A i.e. $x_A$

should be such that $x_A=x_B+gap$ because $A$ was $behind$ B

or $x_A=x_B+80m$ ...(1)

Now as A moves with constant velocity so $x_A=10\times t$

and B is moving with acceleration and initially at rest (u=0) so $x_B=0+a_0{t}^2=a_0{t}^2$

putting above values in equation-1 we get $10t=a_0{t}^2+80$

$\text{If car A catches car B then we should get some value of t i.e. the root of above eqaution should be REAL.}$

So the discriminant should be positive i.e ${10}^2-4\times a_0\times 80\ge 0$ or $a_0\le \frac{5}{16}m/s^2$