Question

The ratio of time taken by two cars P and Q starting from rest moving along a straight road with equal accelerations is $\sqrt{2}:1$, then the :

• A. final velocity of car P > final velocity or car Q.
• B. final velocity of car P < final velocity of car Q.
• C. ratio of $V_P$ to $V_Q$ is 2 : $\sqrt{2}$
• D. ratio of distance travelled by car 'P' to car 'Q' is 2 :1

Answer 1

Answer: (A), (C), (D)

The correct options are
A final velocity of car P > final velocity or car Q.
C ratio of $V_P$ to $V_Q$ is 2 : $\sqrt{2}$
D ratio of distance travelled by car 'P' to car 'Q' is 2 :1

The ratio of time $t_P:t_Q=\sqrt{2}:1$, acceleration is same, Let acceleration is $a$, Initial velocity $u=0$.

Using $v=u+at$ ,

$V_P=0+a{t}_p$ ....1

$V_Q=0+a{t}_Q$ .....2

Subtracting equations 2 from 1,

$V_P-V_Q=a(t_P-t_Q)$= $a(\sqrt{2}-1)$>0 so Option(A) is correct.

Divide equations 1 by 2,

$\frac{V_P}{V_Q}=\frac{\sqrt{2}}{1}$ $\Rightarrow V_P:V_Q=2:\sqrt{2}$ so option (C) is correct.

Using $s=ut+\frac{1}{2}a{t}^2$,

$S_P=\frac{1}{2}a{t}_p^2$ ......3

$S_Q=\frac{1}{2}a{t}_Q^2$ .....4

Divide equation 3 by 4,

$\Rightarrow \frac{S_P}{S_Q}=\frac{t_P^2}{t_Q^2}$ $\Rightarrow S_P:S_Q=2:1$ So option (D) is correct.