Question

Position (Km) - Time (min.) graph is as shown for two cars $A$ and $B$. Both collide at time $t=150$ minute. Then the distance of position $R$ of accident from the starting point $Q$ of car $A$ will be: (Initial distance between the two cars is $500$ km) (Position in the graph shows the distance of the two cars from the point $Q$)

• A. $200km$
• B. $300km$
• C. $250km$
• D. $400km$

Answer 1

Answer: (B)

$300km$

We need to calculate the distance of position 'R' of accident from the starting point 'Q' of car A i.e. the y coordinate of point R in the graph.
It is given that the angle of PR with the y-axis is ${37}^o$
Thus angle of PR with x-axis will be ${127}^o$
Thus slope of line PR is $tan\left({127}^o\right)=-\frac{4}{3}$
Equation of line will be
$y=\frac{-4}{3}x+500\Rightarrow y=\frac{-4}{3}\times 150+500\Rightarrow y=300km$(x=150 min)